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1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 26 1.6. Trigonometric Integrals and Trigonometric Substitutions 1.6.1. TrigonometricIntegrals. Herewediscussintegralsofpow- ers of trigonometric functions. To that end the following half-angle identities will be useful: sin2 x = 1(1 −cos2x), 2 cos2x = 1(1+cos2x). 2 Remember also the identities: 2 2 sin x+cos x = 1, 2 2 sec x = 1+tan x. 1.6.1.1. Integrals of Products of Sines and Cosines. We will study now integrals of the form Z m n sin xcos xdx, including cases in which m = 0 or n = 0, i.e.: Z cosnxdx; Z sinmxdx. The simplest case is when either n = 1 or m = 1, in which case the substitution u = sinx or u = cosx respectively will work. Example: Z sin4xcosxdx = ··· (u = sinx, du = cosxdx) Z 5 5 · · · = u4du = u +C = sin x +C . 5 5 More generally if at least one exponent is odd then we can use the 2 2 identity sin x+cos x = 1totransformtheintegrandintoanexpression containing only one sine or one cosine. 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 27 Example: Z 2 xcos3xdx = Z sin2xcos2xcosxdx sin =Z sin2x(1−sin2x)cosxdx=··· (u = sinx, du = cosxdx) · · · = Z u2 (1 − u2)du = Z (u2 − u4)du u3 u5 = 3 − 5 +C 3 sin5 = sin x − x +C . 3 5 If all the exponents are even then we use the half-angle identities. Example: Z 2 2 Z 1 1 sin xcos xdx = 2(1 −cos2x)2(1+cos2x)dx =1Z 2 2x)dx 4 Z (1 −cos =1 (1−1(1+cos4x))dx 4 2 =1Z (1−cos4x)dx 8 =x−sin4x+C. 8 32 1.6.1.2. Integrals of Secants and Tangents. The integral of tanx can be computed in the following way: Z tanxdx=Z sinx dx=−Z du = −ln|u|+C =−ln|cosx|+C , cosx u where u = cosx. Analogously Z cotxdx = Z cosx dx = Z du = ln|u|+C = ln|sinx|+C , sinx u where u = sinx. 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 28 The integral of secx is a little tricky: Z secxdx = Z secx(tanx+secx)dx = Z secxtanx+sec2xdx= secx+tanx secx+tanx Z du =ln|u|+C = ln|secx+tanx|+C , u 2 where u = secx+tanx, du = (secxtanx+sec x)dx. Analogously: Z cscxdx = −ln|cscx+cotx|+C . More generally an integral of the form Z m n tan xsec xdx can be computed in the following way: (1) If m is odd, use u = secx, du = secxtanxdx. 2 (2) If n is even, use u = tanx, du = sec xdx. Z 3 2 Example: tan xsec xdx = ··· Since in this case m is odd and n is even it does not matter which method we use, so let’s use the first one: (u = secx, du = secxtanxdx) Z 2 Z 2 · · · = tan x secx tanxsecxdx = (u −1)udu | {z }|{z}| {z } u2−1 u du Z = (u3−u)du 4 u2 =u − +C 4 2 = 1sec4x− 1sec2x+C . 4 2 Next let’s solve the same problem using the second method: 1.6. TRIGONOMETRIC INTEGRALS AND TRIG. SUBSTITUTIONS 29 (u = tanx, du = sec2xdx) Z 3 2 Z 3 u4 +C = 1 4 tan x sec xdx = u du = 4 4 tan x+C . | {z }| {z } u3 du Although this answer looks different from the one obtained using the first method it is in fact equivalent to it because they differ in a con- stant: 1 4 1 2 2 1 4 1 2 1 4 tan x = 4(sec x−1) = 4 sec x− 2 sec x+4 . | {z } previous answer 1.6.2. Trigonometric Substitutions. Here we study substitu- tions of the form x = some trigonometric function. Example: Find Z √1−x2dx. Answer: We make x = sint, dx = costdt, hence √ 2 p 2 √ 2 1−x = 1−sin t= cos t=cost, and Z Z √1−x2dx=Z cost costdt 2 = cos tdt =Z 1(1+cos2t)dt (half-angle identity) 2 = t + sin2t +C 2 4 = t + 2sintcost +C (double-angle identity) 2 4 p 2 = t + sint 1−sin t +C 2 2 −1 √ 2 = sin x + x 1−x +C . 2 2 The following substitutions are useful in integrals containing the following expressions:
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