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MATH 109 HANDOUT 1, INTEGRALS OF TRIGONOMETRIC FUNCTIONS Evaluating R sinm x cosn dx. 1. If n is odd, we save one factor of cos x, substitute u = sin x, and express the integrand in terms of u: Z Z sin x cos dx = m sinm x cosn−1 x cos x dx n Z = sinm x(1 − sin2 x)(n−1)/2 cos x dx Z = um (1 − u2 )(n−1)/2 du 2. If m is odd, the same strategy as above works, save one factor of sin x and substitute u = cos x. 3. If both m and n are even, use double-angle formulae and go back to either case 1 or 2. Evaluating R tanm x secn dx. 1. If n is even, save a factor of sec2 x, substitute u = tan x, and express the remaining factors in terms of tan x: Z Z tanm x secn dx = tanm x secn−2 x sec2 x dx Z = tanm x(1 + tan2 x)(n−2)/2 sec2 x dx Z = um (1 + u2 )(n−2)/2 du 2. If m is odd, save a factor of sec x tan x, substitute u = sec x, and express the remaining factors in terms of sec x: Z Z tanm x secn dx = tanm−1 x secn−1 x tan x sec x dx Z = (sec2 x − 1)(m−1)/2 secn−1 xtan x sec x dx Z = Evaluating R (u2 − 1)(m−1)/2 un−1 du sec x dx. This shows in the textbook on top of page 483, following the second approach I gave in the lecture. Let me 1 2 MATH 109 HANDOUT 1, INTEGRALS OF TRIGONOMETRIC FUNCTIONS finish the first approach here: Z Z sec x dx = Z = Z = Z = Z = 1 dx cos x cos x dx cos2 x cos x dx (let u = sin x, then du = cos x dx) 1 − sin2 x du (partial fraction trick, we will learn it) 1 − u2 1 1 1 du + 2 1−u 1+u Z Z 1 1 1 1 = du + du (how to do these two integrals?) 2 1−u 2 1+u 1 1 = − ln|1 − u| + ln|1 + u| + C 2 2 1 1+u |+C = ln| 2 1−u r 1+u = ln +C 1−u r 1 + sin x = ln + C (check the final step) 1 − sin x = ln|sec x + tan x| + C