Integral sqrt x 2 1 x

• [PDF File]Table of Integrals

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  1 2 secxtanx+ 1 2 ln|secxtanx| (76)!secxtanxdx=secx (77)!sec2xtanxdx= 1 2 sec2x (78)!secnxtanxdx= 1 n secnx, n!0 2 (79)!cscxdx=ln|cscx"cotx| (80)!csc2xdx="cotx (81)!csc3xdx=" 1 2 cotxcscx+ 1 2 ln|cscx"cotx| (82)!cscnxcotxdx=" 1 n cscnx, n!0 (83)!secxcscxdx=lntanx TRIGONOMETRIC FUNCTIONS WITH xn (84)!xcosxdx=cosx+xsinx (85)!xcos(ax)dx= 1 a2 ...

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• [PDF File]localhost:8888/nbconvert/html/Documents/TAMUadmin ...

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  Example 1: Compute the indefinite integral of f(x) = x^3 * sqrt(x^2 + 4) Example 2: Compute the integral of f from x=0 to x=2. In [2]: x=symbols('x') # Remember the symbols command allows x to be defined as ju st "x" f=x**3*sqrt(x**2+4) # Recall the ** for exponents. Also notice sqrt for th e square root F=integrate(f,x)

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• [PDF File]Techniques of Integration - Whitman College

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  cos2 x = 1−sin2 x sec2 x = 1+tan2 x tan2 x = sec2 x −1. If your function contains 1−x2, as in the example above, try x = sinu; if it contains 1+x2 try x = tanu; and if it contains x2 − 1, try x = secu. Sometimes you will need to try something a bit different to handle constants other than one. EXAMPLE10.2.2 Evaluate Z p 4− 9x2 dx. We ...

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• [PDF File]Table of Integrals

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  Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=

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• [PDF File]Calculus II, Section6.2, #34 Volumes Set up an integral ...

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  in Figure 1. We know x2+y2 = 1 is the unit circle, but because we are told y ≥ 0, we only use the top half. 1-1 1 x y Figure 1 Beforewe canset up the integralforthe volume ofthe solidofrevolution, weneed to find the coordinates of the points where the curves intersect. We solve x2 = p 1−x2 x4 = 1−x2 x4 +x2 −1 = 0 Let w = x2, then w2 ...

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• [PDF File]Techniques of Integration - Whitman College

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  Then since u = 1− x2: Z x3 p 1− x2 dx = 1 5 (1−x2)− 1 3 (1−x2)3/2 + C. To summarize: if we suspect that a given function is the derivative of another via the chain rule, we let u denote a likely candidate for the inner function, then translate the given function so that it is written entirely in terms of u, with no x remaining in the ...

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• [PDF File]RESIDUE CALCULUS, PART II

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  Integral of the square root round the unit circle Take principal branch : f(z) = ... x2 +1 = π ln2. SUMMATION OF SERIES BY RESIDUE CALCULUS X ...

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• [PDF File]Maxima by Example: Ch.7: Symbolic Integration

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  (%i3) integrate (x/ sqrt (bˆ2 - xˆ2), x); 2 2 (%o3) - sqrt(b - x ) (%i4) diff(%,x); x (%o4) -----2 2 sqrt(b - x ) Example 3 The definite integral can be related to the ”area under a curve” and is the more accessible concept, while the integral is simply a function whose first derivative is the original integrand. Here is a definite ...

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• [PDF File]Integration by substitution

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  The integral becomes Z x=3 x=1 u2 du where we have explicitly written the variable in the limits of integration to emphasise that those limits were on the variable x and not u. We can write these as limits on u using the substitution u = 9+x. Clearly, when x = 1, u = 10, and when x = 3, u = 12. So we require Z u=12

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• [PDF File]Math 104: Improper Integrals (With Solutions)

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  1 (x−1)2/3 dx, if it converges. Solution: We might think just to do Z 3 0 1 (x−1)2/3 dx= h 3(x− 1)1/3 i 3 0, but this is not okay: The function f(x) = 1 (x−1)2/3 is undefined when x= 1, so we need to split the problem into two integrals. Z 3 0 1 (x− 1)2/ 3 dx= Z 1 0 1 (x− 1)2/ dx+ Z 3 1 1 (x− 1)2/3 dx. The two integrals on the ...

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• [PDF File]1 Evaluating an integral with a branch cut

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  1/x−1 when we approach from the lower half plane. Thus the boundary values of f(z) are f(x+i0) = 1 ±ix q 1− 1 x = 1 ±i p x(1−x). Take a curve C going around the interval 0 ≤ x ≤ 1 counterclockwise. We can replace C by such a curve that goes around the interval and stays a distance † from it. Then by taking the limit † → 0 we ...

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• [PDF File]Double integrals

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  1 0 " 3y −xy − y2 2 # y=x y=0 dx = Z 1 0 3x−x2 − x2 2! dx = Z 1 0 3x− 3x2 2! dx = " 3x2 2 − x3 2 # 1 x=0 = 1 Note that Methods 1 and 2 give the same answer. If they don’t it means something is wrong. 0.11 Example Evaluate ZZ D (4x+2)dA where D is the region enclosed by the curves y = x2 and y = 2x. Solution. Again we will carry ...

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• [PDF File]Integral Evaluation - University of Houston

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  Singularities like 1/ x. 2. are non- integrable (even in the PV sense). 22 2 2 2. 1 1. 1 1 1 1 11 sgn( ) ,0,0. b a. x x. dx dx x x a b x x x. ε ε. εε. − + = + + − →∞ > = − < ∫∫. but note that has a PV integral. 6. x. 2. 1. x a. −ε ε. b. Cauchy Principal Value Integrals (cont.)

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• [PDF File]Table of Basic Integrals Basic Forms

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  a 2+ x dx= 1 2 x2 1 2 a2 lnja2 + x2j (13) Z 1 ax2 + bx+ c dx= 2 p 4ac b2 tan 1 2ax+ b p 4ac b2 (14) Z 1 (x+ a)(x+ b) dx= 1 b a ln a+ x b+ x; a6=b (15) Z x (x+ a)2 dx= a a+ x + lnja+ xj (16) Z x ax2 + bx+ c dx= 1 2a lnjax2+bx+cj b a p 4ac 2b2 tan 1 2ax+ b p 4ac b Integrals with Roots (17) Z p x adx= 2 3 (x a)3=2 (18) Z 1 p x a dx= 2 p x a (19) Z ...

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• [PDF File]Definite Integrals by Contour Integration

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  C Z- Z+ Of the poles, only z+ lies inside the unit circle, so I =2πiR+ where R+ is the residue at z+ To find the residue we note that this is a simple pole and if we write the integrand as f(z)=g(z)/h(z) the residue at z+ is: g(z+) h (z 2 2i(az+ +1) 1 i √ 1− a2 Hence the integral required is 2π/ √ 1− a2 Type 2 Integrals Integrals such as I = +∞ −∞ f(x)dx or, equivalently, in ...

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• [PDF File]∫∫ ∫ ∫ ∫∫ - UH

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  9. Set up the integral to find the volume of the solid bounded above by the plane y + z = 1, below by the xy-plane, and on the sides by y=xand x = 4. a. dzdxdy 0 1−y ∫ y2 2 ∫ 0 1 ∫ b. dzdydx 0 1−y ∫ 0 x ∫ 0 1 ∫ c. dzdydx 0 1−y ∫ x 2 ∫

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• [PDF File]FT. SECOND FUNDAMENTAL THEOREM

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  So what (2′) says is: “Let F(x) be an antiderivative for f(x); then F(x) is an antiderivative for f(x) — a true statement, but not a very exciting one (logicians call it a tautology.) The Second Fundamental Theorem (2) looks almost the same as (2 ′ ), but it is actually

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• [PDF File]Integral of x3/x2-1 - MIT OpenCourseWare

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  x3 Integral of x2 −1 Express the integrand as a sum of a polynomial and a proper rational function, then integrate: x3 dx. x2 − 1 Solution The numerator of the integrand has a higher degree than the denominator, so we must use long division to convert the integrand from an “improper fraction” to a “mixed fraction”. x x2 −1 3 − ...

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• [PDF File]GAUSSIAN INTEGRALS

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  the integral by I, we can write I2 = µZ ∞ −∞ e−x2 dx ¶ 2 = Z ∞ −∞ e−x2 dx ∞ −∞ e−y2 dy (2) where the dummy variable y has been substituted for x in the last integral. The product of two integrals can be expressed as a double integral: I2 = Z ∞ −∞ Z ∞ −∞ e−(x2+y2) dxdy The differential dxdy represents an ...

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