ࡱ> Z\STUVWXY7  bjbjUU "7|7|=l::::lN $? _ fe U      B r:k0   CBSE CLASS XII MATHS Integral Calculus Two mark questions with answersQ1.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1.gif" \* MERGEFORMATINET dx Ans1.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1i.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1ii.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1iii.gif" \* MERGEFORMATINET dx = (sec2x + secx tanx) dx = tanx + secx + c Q2.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1iv.gif" \* MERGEFORMATINET Ans2.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1iv.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1v.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1vi.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1vii.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1viii.gif" \* MERGEFORMATINET  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/p.gif" \* MERGEFORMATINET c =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq1viiii.gif" \* MERGEFORMATINET  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/p.gif" \* MERGEFORMATINET c Q3. (3 cotx - 2 tanx)2 dx Ans3. (3 cotx - 2 tanx)2 dx = (9 cot2x + 4 tan2x - 12) dx = [9(cosec2x - 1) + 4(sec2 x - 1) - 12] dx = (4 sec2x + 9 cosec2x - 25) dx = 4 tanx - 9 cotx - 25x + c Q4.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq4.gif" \* MERGEFORMATINET dx Ans4.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq4.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq4i.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq4ii.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq4iii.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq4iv.gif" \* MERGEFORMATINET + c Q5. (1 - sinx) dx Ans5. (1 - sinx) dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq5i.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq5ii.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq5iii.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq5iv.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq5v.gif" \* MERGEFORMATINET Q6. sin2x sin3x dx Ans6. sin2x sin3x dx = (1/2)(2 sin2x sin3x) dx = (1/2) (cosx - cos5x) dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq6.gif" \* MERGEFORMATINET + c Q7. a3x+2 dx, a > 0 Ans7. a3x+2 dx = [a3x+2/3log a] + c Q8.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8.gif" \* MERGEFORMATINET dx Ans8.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8i.gif" \* MERGEFORMATINET dx.........[... elogex = x] =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8ii.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8iii.gif" \* MERGEFORMATINET .dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8iv.gif" \* MERGEFORMATINET + c Q9. cos3x dx Ans9. cos3x dx = (1/4) (cos3x + 3cosx) dx..........[... cos3x = 4cos3x - 3cosx] =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8v.gif" \* MERGEFORMATINET + c Q10.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq10.gif" \* MERGEFORMATINET dx Ans10.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq10.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq8vi.gif" \* MERGEFORMATINET dx = (sec x.tan x - cosec x.cotx) dx = sec x + cosec x + c Q11.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq11.gif" \* MERGEFORMATINET dx Ans11.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq11.gif" \* MERGEFORMATINET dx put x = t2 dx = 2t.dt \ INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq11.gif" \* MERGEFORMATINET  dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq11i.gif" \* MERGEFORMATINET  = 2/(t - 1) dt = 2 log|t - 1| + c = 2log|x - 1| + c Q12. [tannx sec2x]dx Ans12. tannx sec2x dx put tanx = t sec2x.dx = dt ... tannx sec2x.dx = tn.dt =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/q.gif" \* MERGEFORMATINET Q13.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13.gif" \* MERGEFORMATINET Ans13.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13.gif" \* MERGEFORMATINET  Put a2sin2x - b2cos2x = t (a2.2sinx cosx + b2.2sinx cosx) dx = dt (a2 + b2)sin2x dx = dt sin2x dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13i.gif" \* MERGEFORMATINET dt \ INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13ii.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13i.gif" \* MERGEFORMATINET .log |t| =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13i.gif" \* MERGEFORMATINET .log|a2 sin2x - b2 cos2x| + c Q14.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13iii.gif" \* MERGEFORMATINET Ans14.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq13iii.gif" \* MERGEFORMATINET =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq14.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq14iv.gif" \* MERGEFORMATINET +  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq14i.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq14ii.gif" \* MERGEFORMATINET - ( INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq14iii.gif" \* MERGEFORMATINET ) + c = log |x + 1| - [ INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq14iii.gif" \* MERGEFORMATINET ] + c Q15.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq15.gif" \* MERGEFORMATINET .dx Ans15.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq15.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq15i.gif" \* MERGEFORMATINET dx =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq16i.gif" \* MERGEFORMATINET  Let cos x = t2 - sin x dx = 2t. dt sin x dx = -2t. dt \ INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq16i.gif" \* MERGEFORMATINET  = -2  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq16ii.gif" \* MERGEFORMATINET  = 2 (t6 - t2) dt =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq16iii.gif" \* MERGEFORMATINET + c = 2  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aq16iv.gif" \* MERGEFORMATINET + c Q16. secq/tan2q dq Ans16. secq/tan2q dq =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq16.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aa16i.gif" \* MERGEFORMATINET  Let sin q = t cosq dq = dt \  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aa16i.gif" \* MERGEFORMATINET =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq16i.gif" \* MERGEFORMATINET  = -1/t + c = -1/sin q + c = - cosec q + c Q17.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq17.gif" \* MERGEFORMATINET Ans17.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq17.gif" \* MERGEFORMATINET  Put x = t2 dx = 2t.dt \  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq17.gif" \* MERGEFORMATINET =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq17i.gif" \* MERGEFORMATINET  = 2cost.dt = 2sin t = 2sin x + c Q18.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq18.gif" \* MERGEFORMATINET Ans18.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq18.gif" \* MERGEFORMATINET  Put xex = t (xex + ex) dx = dt ex (x + 1) dx = dt  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq18.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq18i.gif" \* MERGEFORMATINET = sec2t dt = tant + c = tan (xex) + c Q19.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq19.gif" \* MERGEFORMATINET Ans19.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq19.gif" \* MERGEFORMATINET  Let 1 + log x = t (1/x).dx = dt \  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq19.gif" \* MERGEFORMATINET  = t2.dt = (t3/3) + c =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq19i.gif" \* MERGEFORMATINET + c Q20.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq20.gif" \* MERGEFORMATINET Ans20.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq20.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq20i.gif" \* MERGEFORMATINET  Put ex = t ex.dx = dt \  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq20.gif" \* MERGEFORMATINET  =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq20ii.gif" \* MERGEFORMATINET  = sin-1t + c = sin-1(ex) + c Q21. logx dx Ans21. logx dx = logx dx -  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/aaq22x.gif" \* MERGEFORMATINET  = x logx - (1/x) . x dx = x logx - x + c = x(logx - 1) + c Q22. cos x dx Ans22. cos x dx put x = t2 dx = 2t dt \ cos x dx = 2t cost dt = 2(tsint - 1. sint dt) = 2(tsint + cost) + c = 2(x sin x + cos x) + c Four mark questions with answersQ1. tanx tan2x tan3x dx Ans1. We have tan3x = tan(2x + x) [(tan2x + tanx)/(1 - tan2x tanx)] tan3x - tan3x tan2x tanx = tan2x + tanx [after cross multiplying] tan3x tan2x tanx = tan3x - tan2x - tanx \ tanx tan2x tan3x dx = (tan3x - tan2x - tanx) dx = (1/3) log|sec3x| - (1/2) log|sec2x| - log|secx| + c Q2.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/4q2i.gif" \* MERGEFORMATINET Ans2.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/4q2ii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/4q2iii.gif" \* MERGEFORMATINET  Put cos a + cotx sina = t2 = -cosec2x sina = 2t dt cosec2x dx = (-2/sina)t dt  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/4q2iv.gif" \* MERGEFORMATINET  = (-2/sina) (t dt)/t = (-2/sina)t + c = -2/sin a (cota + cotx sina)1/2 + c Q3. cos5x (sinx) dx Ans3. cos5x (sinx) dx = cos4x . cosx . (sinx). dx = (1 - sin2x)2 . cosx (sinx) dx put sinx = t2 \ cosx dx = 2t dt \ (1 - sin2x)2 cosx (sinx) dx = (1 - t4)2.t.2t dt = 2(1 + t8 - 2t4)t2 dt = 2(t10 - 2 t6 + t2)dt = 2(t11/11) - 2(t7/7) + 2(t3/3) + c = 2[{(sinx)11/2/11} - 2(sinx)7/2/7 +2 (sinx)3/2/3] + c Q4. e5x cos4x dx Ans4. I = e5x cos4x dx [Integrating by parts we get] = (1/4)(e5x sin4x) - 5/4e5x sin4x dx = 1/4(e5x sin4x) - 5/4[-1/4 e5x cos4x + 5/4e5x cos4x dx] = 1/4(e5x sin4x) + 5/16(e5x cos4x) - (25/16) I + c 4(1/16) I = (1/4) e5x(5/4cos 4x + sin 4x) + c I = (4/41)e5x[cos 4x +(5/4) sin 4x] + c Q5. cos(logx).dx Ans5. I = cos(logx) dx = x cosx (logx) + [{sin(logx)}/x] x dx = x cosx log x + x sinx log x - [{cos(logx)}/x] x dx = x cosx (log x) + x sinx (log x) - I + c 2I = x[cos (logx) + sin (logx)] + c I = x/2[cosx (logx) + sin (logx)] + c Q6.log[(1 + x) + (1 - x)]dx Ans6. log[(1 + x) + (1 - x)]dx = x log[(1 + x) + (1 - x)] - d/dx{ log[(1 + x) + (1 - x)]} . x dx = x log[(1 + x) + (1 - x)] -[1/(1 + x) + (1 - x)].[1/{2(1 + x)} - 1/{2(1-x)]}.x dx =x log[(1 + x) + (1 - x)] -1/2[1/{(1 + x) + (1 - x)}].[{(1 - x) - (1 + x)}/(1 - x)2].x dx =x log[(1 + x) + (1 - x)] -1/2[{(1 - x) - (1 + x)}2/{(-2x)(1 - x2)}].xdx = x log [(1 + x) + (1 - x)] + 1/4[{2 - 2(1 - x2)}/(1 - x2)]dx = x log [(1 + x) + (1 - x)] + (1/4) . 2[{1/(1 - x2)} -1]dx = x log [(1 + x) + (1 - x)] + (1/2) . (sin-1x - x) + c Q7. x3tan-1x dx Ans7. x3 tan-1x dx = x4/4 tan-1x - {1/(1 + x2)}.x4/4 dx = x4/4 tan-1x - (1/4) [(x4 - 1 + 1)/(x2 + 1)]dx = x4/4 tan-1x - 1/4[(x2 + 1)(x2 - 1)/(x2 + 1) + 1/(x2 + 1)]dx = x4/4 tan-1x - 1/4(x3/3 - x + tan-1x) + c Q8. [(x + x2/3 + x1/6)/x(1 + x1/3)]dx Ans8. I = [(x + x2/3 + x1/6)/x(1 + x1/3)] dx Put x = t6...................[Here t6 has been taken ... 6 is L.C.M. of denominators of fractional powers of x] dx = 6t5 dt \ I = [{(t6 + t4 + t).6t5}/{t6(1 + t2)}]dt = 6[(t5 + t3 + 1)/(t2 + 1)]dt = 6[t3 + {1/(t2 + 1)}]dt = 6(t4/4 + tan-1t) + c = (3/2)x2/3 + tan-1x1/6 + c Q9. [(x1/2 + x1/3)/(x5/4 - x7/6)]dx Ans9.[(x1/2 + x1/3)/(x5/4 - x7/6)]dx put x = t12 dx = 12t11 [Here t12 has been taken ... 12 is L.C.M. of denominators of fractional powers of x] \ [{(t6 + t4).12t11}/(t15 - t14)]dt = 12[t4 (t2 + 1) t11/t14(t - 1)]dt =12[{t (t2 + 1)}/(t - 1)]dt = 12[t3/(t - 1)dt + 12[t/(t - 1)]dt = 12[(t3 -1 + 1)/(t - 1)]dt + 12[(t - 1 + 1)/(t - 1)]dt = 12[(t2 + t + 1)dt + [2/(t - 1)]dt +12 dt = 12[t3/3 + t2/2 + 2t + 2log|t - 1|] + c = 12[x1/4/3 + x1/6/2 + 2x1/12 + 2 log|x1/12 - 1|] + c Q10. dx/[(x - 1)3/4. (x + 2)5 ] Ans10. dx/ [(x - 1)3/4. (x + 2)5] = dx/[(x + 2)(x - 1){(x + 2)/(x - 1)}1/4] put {(x + 2)/(x - 1)} = t4 x = {(t4 + 2)/(t4 - 1)} \ dx = [-12t3/(t4 - 1)2]dt \ I = -[12t3/(t4 - 1)2. {3/(t4 - 1)} . {3t4/(t4 - 1).t]dt = -(12/9)(t3/t5)dt = -(4/3) dt/t2 = (4/3)t + c = (4/3)[(x - 1)/(x + 2)]1/4 + c Q11. tan32x sec2x dx Ans11. tan32x sec2x dx = tan22x .sec2x .tan2x. dx = (sec22x - 1) sec2x tan2x dx = let sec2x = t 2 sec2x tan2x dx = dt sec2x tan2x dx = 1/2 dt \ I = (1/2)(t2 - 1)dt = (1/2)(t3/3 - t) + c = (1/2)[sec3(2x/3) - sec2x] + c Q12. [(cos5x + cos4x)/(1 - 2cos3x)]dx Ans12. I = [(cos5x + cos4x)/(1 - 2cos3x)]dx = [(cos5x + cos4x)sin3x/(sin3x - 2sin3x cos3x)]dx = [{(2sin 3x/2 cos3x/2).(2cos 9x/2. cos x/2)}/(sin3x - sin6x)]dx = [(2sin 3x/2 cos3x/2)(2cos9x/2. cosx/2)]/[-2 cos(9x/2).sin(3x/2)]dx = - 2cos3x/2.cosx/2 dx = - (cos2x + cosx) dx = -[(sin2x)/2 + sinx] + c Q13. [(x sin-1 x)/(1 - x2)]dx Ans13. [(x sin-1 x)/(1 - x2 )]dx Put x = sin q dx = cosq dq \I = [{sinq sin-1(sinq).cosq)}/(1 - sin2q)]dq = q sinq dq = -q cos q - 1(-cosq)dq = -q cosq + sinq + c = x - (1 - x2).sin-1x + c Q14. (x - 2)[(x + 3)/(x - 3)]dx Ans14. (x - 2)[(x + 3)/(x - 3)]dx = [(x - 2)(x + 3)/(x - 3)(x + 3)]dx =[(x2 + x - 6)/(x2 - 9)]dx = [(x2 - 9 + x + 3)/(x2 - 9)]dx = (x2 - 9)dx + x/(x2 - 9)dx + 3/(x2 - 9)dx = [x2- (3)2]dx + 1/2 2x/(x2 - 9)dx + 3 log|x + (x2 - 9)| + c = [{x(x2 - 9)}/2] - 9/2 log|x + (x2 - 9)| + (x2 - 9) +3 log|x+(x2 - 9)|+ c = {x(x2 - 9)}/2 - 3/2 log|x + (x2 - 9)| + (x2 - 9) + c Q15. (sinx/sin3x)dx Ans15. (sinx/sin3x)dx = {sinx/(3sinx - 4sin3x)}dx = 1/(3 - 4 sin2x)dx I = [sec2x/(3sec2x - 4tan2x)]dx (Divide numerator and denominator by cos2x) = [sec2x/(3 - tan2x)]dx Put tanx = t sec2x dx = dt \ I = dt/[(3)2 - t2] = 1/(23) log |(3 + t)/(3 - t)| + c = 1/(23) log |(3 + tanx)/(3 - tanx)| + c Q16. [(2sin2x - cosx)/(6 - cos2x - 4sinx)]dx Ans16. I = [(2sin 2x - cosx)/(6 - cos2x - 4sinx)]dx = [(2.2sinx cosx - cosx)/[6 - (1 - sin2x) - 4 sinx]dx = [{(4sinx - 1) cosx}/(sin2x - 4sin x + 5)]dx put sinx = t cosx dx = dt \ I = [{(4t - 1)}/(t2 - 4t + 5)]dt = {2(2t - 4) + 7}/(t2 - 4t + 5) dt = 2 (2t - 4)/(t2 - 4t + 5) dt + {(7)/(t2 - 4t + 5)} dt = 2log |t2 - 4t + 5| + 7/{(t - 2)2 + 1}.dt = 2log |t2 - 4t + 5| + 7 tan-1 (t - 2) + c = 2log |sin2x - 4sinx + 5| + 7tan-1(sinx - 2) + c Q17. [(2x + 1)/(x + 1)(x - 2)]dx Ans17. [(2x + 1)/(x +1)(x - 2)]dx Let [(2x + 1)/(x + 1)(x - 2)]dx = (A)/(x + 1) + (B)/(x - 2) 2x + 1 = A(x - 2) + B(x + 1) ....................... (i) put x = 2, -1 in (i) we get B = 5/3, A = 1/3 \ [(2x + 1)/(x + 1)(x - 2)]dx = (1/3)1/(x + 1)dx + (5/3)1/(x - 2)dx = (1/3)log|x + 1| + (5/3)log|x - 2| + c Q18. dx/[(2 - x)(x2 + 3)] Ans18. dx/(2 - x)(x2 + 3) Let 1/(2 - x)(x2 + 3) = A/(2 - x) + (Bx + c)/(x2 + 3) 1 = A(x2 + 3) + (Bx + C)(2 - x)........................ (i) put x = 2 in (i), \ A = 1/7 equating coefficient of x2 and x on both the sides of (i) \A - B = 0 and 2B - C = 0 = A = B = 1/7 and C = 2/7 \ dx/[(2 - x)(x2 + 3)] = 1/7[1/(2 - x)]dx + 1/7[(x + 2)/(x2 + 3)]dx = -(1/7) log|2 - x| + (1/7 x 1/2) 2x/(x2 + 3) + 2/(x2 + 3) dx = -(1/7) log |2 - x| + (1/14) log|x2 + 3| + (2/3)tan-1(x/3) + cQ19. (tanx + cotx)dx Ans19. (tanx + cotx)dx = [(sinx/cosx) + cosx/sinx)]dx = (sinx + cosx)/(sinxcosx). Let sinx - cosx = t 1 - 2sinx cosx = t2 or sinx cosx = (1 - t2)/2 (cosx + sinx) dx = dt \ I = {dt/[(1 - t2)/2]} = [ (2)/(1 - t2)]dt = 2 sin-1t + C = 2 sin-1(sinx - cosx) + CQ20. dx/(sin4x + cos4x) Ans20.dx/(sin4x + cos4x) (Divide numerator and denominator by cos4x) sec4x /(tan4x + 1)dx = (sec2x.sec2x)/[(tan2x)2 + 1]dx =(1 + tan2x) sec2x /[(tan2x)2 + 1]dx Let tanx = t so, sec2x dx = dt \ I = [(1 + t2)/(t4 + 1)]dt = [1 + (1/t2)]/{t2 + (1/t2) - 2 + 2}dt = [1 + (1/t2)]/[{t - (1/t)}2 + 2]dt Put [t - (1/t)] = z so, {1 + (1/t2)}dt = dz Therefore, I =dz/{z2 + (2)}2 = (1/2) tan-1(z/2) + c = (1/2) tan-1 [(t2 - 1)/2t] + c = (1/2) tan-1[(tan2x - 1)/(2 tanx)] + C Six mark questions with answersQ1. Sketch the graph y = |x + 1|. Evaluate  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q1i.gif" \* MERGEFORMATINET |x + 1|dx. What does the value of this integral represent on this graph? Ans1.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q1v.gif" \* MERGEFORMATINET  y = |x + 1| = {(x + 1) if x > - 1} and {- (x + 1) if x < -1} Graph is a pair of lines through x = -1 making angle 45o & 135o with x-axis because slope of the lines are + 1. Further to draw the lines we need two points on each lines. Join the pair to point (-1, 0) and (0, 1) by line to get graph to y = x + 1. Also join the pair of points (-1, 0) and (-2, 1) by line to get graph of y = -(x + 1). Evaluation of integral.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q1ii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q1iii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q1iv.gif" \* MERGEFORMATINET  = -[-1/2 - 4] + [4 + 1/2] = 1/2 + 4 + 4 + 1/2 = 9 Representation of the value 9 of integral on graph.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q1i.gif" \* MERGEFORMATINET |x + 1|dx = 9 represent the area bounded by the curve y = |x + 1|, X-axis and the lines x = -4 and x = 2 i.e. it is equal to the sum of the area of triangle ABD and ACE. = (1/2)(9) + (1/2)(9) = 9/2 + 9/2 = 9 [... Area = 1/2 base x height] Q2. Find the area bounded by x = at2, y = 2 at between ordinates corresponding to t = 1 and t = 2. Ans2. Eliminate 't' form give two relation y = 2at t = y/2a x = at2 = a(y/2a)2 = y2/4a y2 = 4ax ..............(1) Also t = 1 x = a and y = 2a t = 2 x = 4a and y = 4a Now (1) is parabola which is symmetrical about x-axis  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q2i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q2ii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q2iii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q2iv.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q2v.gif" \* MERGEFORMATINET Q3. Compute the area bounded by the lines x + 2y = 2, y - x = 1 & 2x + y = 7. Ans3.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3xii.gif" \* MERGEFORMATINET  Given lines are x + 2y.......(1), y - x = 1 ...........(2), 2x + y = 7 ........(3) Line (1), meets x-axis at (2, 0) and y-axis at (0, 1) Join (2, 0) and (0, 1) to get the graph of first lines. Similarly second lines is join of (-1, 0) and (0, 1) and third line is join (0, 7) and (7/2, 0) Lines (1) and (2) meet in (0, 1) Lines (2) and (3) meet in (2, 3) Lines (3) and (1) meet in (4, -1). Thus the points of intersection of the three lines are A(0,1), B(2,3), C(4, -1). Area bounded by the lines is shown shaded in the figure. Required area = Area ABC = Area ABD + Area DEB + Area DLE + Area LCE} .........(4) Now Area ADB = Area DBAO - Area DAO  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3ii.gif" \* MERGEFORMATINET  = 4 - 1/2(2) = 3 ........(5) Area DEB =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3iii.gif" \* MERGEFORMATINET (7 - 2x)dx  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3iv.gif" \* MERGEFORMATINET  = (49/2 - 49/4) - (14 - 4) = 49/4 - 10 = 9/4 ...........(6)  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3v.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3vi.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3vii.gif" \* MERGEFORMATINET  Area LCE = Area ELCM - Area ECM  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3viii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3ix.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3x.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q3xi.gif" \* MERGEFORMATINET  = Modulus (-3/16) = 3/16......(8) Using the values of areas obtained (5), (6), (7), (8) in (4) we have Required area = 3 + 9/4 + 9/16 + 3/16 = 96/16 = 6 sq. units Q4. Sketch the graph y = (x) + 1 in {0, 4} and determine the area of the region enclosed by the curve, the axis of x and the lines x = 0, x = 4. Ans4.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q4iii.gif" \* MERGEFORMATINET  y = (x) + 1 x = y - 1 x = (y - 1)2.........(1) Shift origin to the point (0, 1) \ x = X + 0 and y = Y + 1 put (1) we get X = Y2 which is parabola (Right Handed) with vertex at (0,1). Plot these points and join them by free hand curve to get rough sketch Required Area (Shaded)  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q4i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q4ii.gif" \* MERGEFORMATINET  = 2/3 (8) + 4 = 16/3 + 4 = 28/3 sq.units Q5. Find the area of the region {(x, y) : y2 4x, 4x2 + 4y2 9} Ans5.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6qv.gif" \* MERGEFORMATINET  Let R = {(x, y):y2 4x, 4x2 + 4y2 9} = {x, y:y2 4x} {(x, y):4x2 + 4y2 < 9} = R1 R2 Where R1 = {x, y:y2 4x} represent the region inside the parabola y2 = 4x with vertex (0, 0) and x-axis as it axis. and R2 = {x , y:4x2 + 4y2 < 9} represent the interior of the circle 4x2 + 4y2 = 9 with centre (0, 0) and radius (3/2). Thus the region R which is intersection of R1 & R2 is shown shaded in the figure. Now to find the points intersection of the given curves, we solve their equation y2 = 4x ...........(1) and 4x2 + 4y2 = 9 ..........(2) simultaneously. Using (1) & (2) we get 4x2 + 4(4x) = 9 4x2 + 16x - 9 = 0 4x2 + 18x - 2x - 9 = 0 (2x - 1) (2x + 9) = 0 x = 1/2, -9/2 From (1) x = 1/2 y = + 2 and x = -9/2 y = imaginary quantity Thus the points of intersection of (1) and (2) are A(1/2, 2) and B (1/2, -2) Also both the curve are symmetrical about x-axis \ Required Area = Area OBPAO - 2.Area (OPAO) = 2[Area ODA + Area DPA]  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q5i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q5ii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q5iii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q5iv.gif" \* MERGEFORMATINET  = 8/3(1/22) + (9/4)p/2 - (1/2) - 9/4 sin-1 (1/3) = (22/3) - 1/2 + 9p/8 - 9/4 sin-1 (1/3) sq.units Q6. Find the area of the region enclosed between the two circle x2 + y2 = 1 and (x - 1)2 + y2 = 1 Ans6.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q6iv.gif" \* MERGEFORMATINET  x2 + y2 = 1............(1) (x - 1)+ y2 = 1.........(2) Intersect at the point obtained by solving (1) and (2) From (1) y2 = 1 - x2, putting in (2) we get (x - 1)2 + 1 - x2 = 1 -2x + 1 = 0 x = 1/2 From (1) x = 1/2 y = + 3/2 \ (1) and (2) intersect at A[1/2, 3/2] and B [1/2, -3/2] Center of first circle is (0, 0) and radius = 1 Also centre of second circle is (1, 0) and radius = 1 Also both, the circle are symmetrical about x-axis \ Required area is shown shaded \ Required area = Area OACB = 2(Area OAC) = 2[Area OAD + Area DCA]  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q6i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q6ii.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q6iii.gif" \* MERGEFORMATINET  = -3/4 - p/6 - (-p/2) + p/2 - (+ 3/4) - p/6 = -3/4 - p/6 + p/2 + p/2 - 3/4 - p/6 = (2p/3 - 3/2) sq. units Q7. Find the area of the region {(x, y)}:x2 + y2 1 x + y}. Ans7.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q7iii.gif" \* MERGEFORMATINET  The required area is the area between the circle x2 + y2 = 1 and line x + y = 1. Circle (1) has centre (0, 0) and radius 1 and line to meets x-axis at A (1, 0) y-axis at B (0, 1). The circle (1) also passes through A and B. Hence points of intersection of (1) and (2) are A (1, 0) and B (0, 1)  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q7i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q7ii.gif" \* MERGEFORMATINET  = (1/2) sin-1(1) - 1/2 = (1/2)(p/2) - 1/2 = (p/4 - 1/2) sq.units Q8. Using integration find the area of DABC with vertices at A (2, 5) B (4, 7) and C (6, 2). Ans8.  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q8iii.gif" \* MERGEFORMATINET  Equation the AB of D ABC is (y - 5) = [{(7 - 5)/(4 - 2)}(x - 2)] y = x + 3 Similarly equation of sides BC and CA are 5x + 2y - 34 = 0 and 3x + 4y - 26 = 0 Required Area = Area of D ABC = Area LMBA + Area of MNCB - Area LNCA  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q8i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q8ii.gif" \* MERGEFORMATINET  = [(8 + 12) - (2 + 6)] + [(102 - 45) - (68 - 20)] - [(39 - 27/2) - (13 - 3/2)] = [20 - 8] + (57 - 48) - [51/2 - 23/2] = 12 + 9 - 14 = 7 sq.units Q9. x-11(1 + x4)-1/2 dx Ans9. x-11(1 + x4)-1/2 dx = 1/x11(1 + x4)1/2dx I = 1/x13(1/x4 + 1)1/2 dx = (1/x8) x5(1 + 1/x4)1/2 dx put 1 + 1/x4 = t2 1/x4 = t2 - 1 -4/x5 dx = 2tdt 1/x5 dx = -1/2 tdt \ I = (-1/2) (t2 - 1)2t/t dt = (-1/2) (t4 - 2t2 + 1)dt = (-1/2) (t5/5 - 2 t3/3 + t) + c = (-1/2)[1/5(1 + 1/x4)5/2 - 2/3 (1 + 1/x4)3/2 + (1 + 1/x4)1/2 ] + c Q10. Evaluate  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q10i.gif" \* MERGEFORMATINET . Ans10. I =  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q10i.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q10ii.gif" \* MERGEFORMATINET  Put sin x = t cos x dx = dt x = 0 t = 0 and x = p/4 t = 1/2  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q10iii.gif" \* MERGEFORMATINET [after solving through partial fractions]  INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q10iv.gif" \* MERGEFORMATINET   INCLUDEPICTURE "http://www.vidyavahini.ernet.in/shishya/products/AcademicContent/CBSE/XII/Maths/xii%20maths%20integral%20calculus/images/6q10v.gif" \* MERGEFORMATINET  &'*IJKMOP[\]^cd  xkjOCJOJQJUaJjCJOJQJUaJj 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