Int dx x tan x 1

• [PDF File]8.6 Integrals of Trigonometric Functions - Contemporary Calculus

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  8.6 Integrals of Trigonometric Functions Contemporary Calculus 5 If the exponent of secant is even, factor off sec2(x), replace the other even powers (if any) of secant using sec2(x) = tan2(x) + 1, and make the change of variable u = tan(x) (then du = sec2(x) dx ). If the exponent of tangent is odd, factor off sec(x)tan(x), replace the remaining even powers (if any) of

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• [PDF File]Table of Integrals - Rice University

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  dx= 1 a tan 1 x a (9) Z x a 2+ x dx= 1 2 lnja2 + x2j (10) Z x 2 a 2+ x dx= x atan 1 x a (11) Z x3 a 2+ x dx= 1 2 x2 1 2 a2 lnja2 + x2j (12) Z 1 ax2 + bx+ c dx= 2 p 4ac b2 tan 1 p 2ax+ b 4ac b2 (13) Z 1 (x+ a)(x+ b) dx= 1 b a ln a+ x b+ x; a6= b (14) Z x (x+ a)2 dx= a a+ x + lnja+ xj (15) Z x ax2 + bx+ c dx= 1 2a lnjax2 + bx+ cj b a p 4ac b2 tan ...

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

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  1 e−x2 dx, (b) Z ∞ 1 sin2(x) x2 dx. Solution: Both integrals converge. (a) Note that 0 < e−x2 ≤ e−x for all x≥ 1, and from example 1 we see R∞ 1 e−x dx= 1 e, so R∞ 1 e−x2 dx converges. (b) 0 ≤ sin2(x) ≤ 1 for all x, so 0 ≤ sin2(x) x 2 ≤ 1 x for all x≥ 1. Since R∞ 1 1 x2 dx converges (by p-test), so does R∞ 1 ...

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• [PDF File]FORMULAS TO KNOW - Brown University

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  FORMULAS TO KNOW Some trig identities: sin2x+cos2x = 1 tan2x+1 = sec2x sin 2x = 2 sin x cos x cos 2x = 2 cos2x 1 tan x = sin x cos x sec x = 1 cos x cot x = cos x sin x csc x = 1 sin x Some integration formulas:

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• [PDF File]7.1 Basic Integration - University of Utah

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  2sin x= 1−cos2x 2 and 2cos x= 1+cos2x 2. B. Form ∫sinm x cosn xdx: If m or n is odd, use Pythagorean identity. If both m and n are even, use half-angle identities. C. Form ∫sin(mx) cos(nx)dx or ∫sin(mx) sin(nx)dx or ∫cos(mx) cos(nx)dx: Use product identities. sin(mx)cos(nx)= 1 2 (sin((m+n)x)+sin((m−n)x)) sin(mx)sin(nx)= −1 2 (cos ...

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• [PDF File]University of South Carolina

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  tan2x + 2 tan x +2 x3 + x2 dx x4 + 2x2 1 3 + cos 0 2 — cos O 2 + 2 cos O + sin 0 1 + sin x sec x Sin X dx sin 2x 99. 101. Find the area of the surface generated by revolving the curve y = cosh x, 0 x 1, around the x-axis. 102. Find the length of the curve y = e 103. (a) Find the area Ab of the surface generated by re-

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• [PDF File]Table of Basic Integrals Basic Forms

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  dx= 1 a tan 1 x a 1 (10) Z x a2 + x2 dx= 1 2 lnja2 + x2j (11) Z x2 a 2+ x dx= x atan 1 x a (12) Z x3 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 ...

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• [PDF File]CALCULUS TRIGONOMETRIC DERIVATIVES AND INTEGRALS - CSUSM

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  CALCULUS TRIGONOMETRIC DERIVATIVES AND INTEGRALS STRATEGY FOR EVALUATING R sinm(x)cosn(x)dx (a) If the power n of cosine is odd (n =2k +1), save one cosine factor and use cos2(x)=1sin2(x)to express the rest of the factors in terms of sine:

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• [PDF File]Trigonometric Integrals - CMU

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  1 n 1 tann 1(x) Z tann 2(x) dx Notice, now we have reduce the problem to an easier problem, since the power of tan is reduced by two. Eventually, by subtracting 2 over and over again, we are either integrating tan(x) or tan2(x). In fact, we can even use the reduction rule on tan2(x) and reduce it to tan0(x) = 1. Example 3. Z tan6(x) dx

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• [PDF File]Table of Integrals

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  1 + x2 dx= tan 1 x (8) Z 1 a2 + x2 dx= 1 a tan 1 x a (9) Z x a 2+ x dx= 1 2 lnja2 + x2j (10) Z x2 a 2+ x dx= x atan 1 x a (11) Z x3 a 2+ x dx= 1 2 x2 1 2 a2 lnja2 + x2j (12) Z 1 ax2 + bx+ c dx= 2 p 4ac b2 tan 1 p 2ax+ b 4ac b2 (13) Z 1 (x+ a)(x+ b) dx= 1 b a ln a+ x b+ x; a6= b (14) Z x (x+ a)2 dx= a a+ x + lnja+ xj (15) Z x ax2 + bx+ c dx= 1 ...

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• [PDF File]Integration by substitution

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  3. Finding Z f(g(x))g′(x)dx by substituting u = g(x) Example Suppose now we wish to find the integral Z 2x √ 1+x2 dx (3) In this example we make the substitution u = 1+x2, in order to simplify the square-root term. We shall see that the rest of the integrand, 2xdx, will be taken care of automatically in the

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• [PDF File]Int(dx)/(x cos^(2)(1 log x))

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  Int(dx)/(x cos^(2)(1 log x)) Int(dx)/(x cos^(2)(1+log x)). Int log(1+cos x)-x tan((x)/(2)) dx. Int_(-pi/2)^( pi/2)cos x log((1+x)/(1-x))dx is equal to.

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• [PDF File]u aga - Lamar University

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  bx a x tan sec 122 222a tan b abx x sec 1 tan22 Ex. 16 2 16 x 49x dx 22 x 33sin cos dx d 49 x2 4 4sin 4cos 2cos22 Recall x2 x. Because we have an indefinite integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d

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• [PDF File]Q: Evaluate the indefinite integral (tan x) d x and the definite ...

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  For x < 0 and x > /2, tan x is negative and the square root complex. There is a singularity at x = /2. Writing x = /2 – y, we have tan x = cos y / sin y = 1/tan y, which is close to 1/y for y small. Thus the singularity in t(tan x) diverges as 1/t( /2 – x). We also see from this symmetry that ∫ =∫ / 2 / 4 / 4 0 tan tan π π π x dx

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• [PDF File]Integration using trig identities or a trig substitution

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  1 a2 +x2 dx: We make the substitution x = atanθ, dx = asec2 θdθ. The integral becomes Z 1 a2 +a2 tan2 θ asec2 θdθ and using the identity 1+tan2 θ = sec2 θ this reduces to 1 a Z 1dθ = 1 a θ +c = 1 a tan−1 x a +c This is a standard result which you should be aware of and be prepared to look up when necessary. Key Point Z 1 1+x2 dx ...

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• [PDF File]Table of Integrals - Oregon State University

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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]Section 7.3, Some Trigonometric Integrals - University of Utah

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  1 2 Z ˇ ˇ cos(2m)x 1 dx = 1 2 1 2m sin(2m)x x ˇ ˇ = ˇ 4 Integrals of the form R tan nxdx and R cot xdx In the tangent case, we will use tan 2x= sec x 1. In the cotangent case, we will use cot2 x= csc2 x 1. Here, we will only replace tan 2xor cot x, distribute, integrate what we can, then repeat as necessary. Examples 1.Find R tan4 xdx Z ...

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