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Week 13 – Trigonometric Integrals Powers of Sines and Cosines To evaluate trigonometric integrals of the type Z sinn x cosm x dx • If n is odd, use the substitution: u = cos x, du = − sin x dx and then the identity cos2 x + sin2 x = 1 if m is odd, try: u = sin x, du = cos x dx If both are odd, it doesn’t matter. • If both n and m are even, use the identities sin2 x = 1 − cos 2x 2 1 + cos 2x 2 to simplify the integral. cos2 x = Z sin3 x cos2 x dx Exercise 13-1: Evaluate Solution: Using u = cos x, du = − sin x dx we obtain: Z 3 Z 2 sin2 x cos2 x sin x dx sin x cos x dx = Z = (1 − u2 )u2 (−du) Z = u4 − u2 du = u5 u3 − +c 5 3 = cos5 x cos3 x − +c 5 3 Z Exercise 13-2: Evaluate sin2 x cos2 x dx Solution: Z 1 − cos 2x 1 + cos 2x · dx 2 2 1 = 4 Z 1 = 4 Z 1 = 8 Z = 1 − cos2 2x dx 1− 1 + cos 4x dx 2 1 − cos 4x dx x sin 4x − +c 8 32 Z Exercise 13-3: Evaluate cos7 x dx Solution: Z Z 7 cos x dx = cos6 x · cos x dx Z = Z = Z = (1 − sin2 x)3 · cos x dx (1 − u2 )3 du u = sin x (1 − 3u2 + 3u4 − u6 ) du 3 1 = u − u3 + u5 − u7 + c 5 7 3 1 = sin x − sin3 x + sin5 x − sin7 x + c 5 7 Exercise 13-4: Evaluate the following integrals: Z a) sin6 x cos3 x dx Z b) Z c) Z d) Z e) sin5 x cos5 x dx sin8 x cos x dx cos4 x dx sin4 x cos4 x dx Products of Sines and Cosines Adding the identities sin(A + B) = sin A cos B + cos A sin B sin(A − B) = sin A cos B − cos A sin B we obtain: sin(A + B) + sin(A − B) = 2 sin A cos B In other words 1 sin(A + B) + sin(A − B) 2 Similarly, starting with sin A cos B = cos(A + B) = cos A cos B − sin A sin B cos(A − B) = cos A cos B + sin A sin B we obtain: 1 cos A cos B = cos(A + B) + cos(A − B) 2 1 sin A sin B = cos(A − B) − cos(A + B) 2 Z Exercise 13-5: Evaluate sin 7x cos 8x dx Solution: Using the above identities, we obtain: Z Z 1 sin 7x cos 8x dx = sin 15x − sin x dx 2 = − cos 15x cos x + +c 30 2 Exercise 13-6: Evaluate the following integrals: Z a) sin 5x sin 3x dx Z b) sin 4x cos 8x dx Z c) cos 3x cos x dx 2 Integrals containing secant and tangent Remember the formulas: d tan x = sec2 x dx d sec x = sec x tan x dx and sec2 x = 1 + tan2 x Z Exercise 13-7: Evaluate sec x dx Solution: Using the derivative formulas for sec x and tan x, we obtain: d (sec x + tan x) = sec x (sec x + tan x) dx Z Z sec x dx = sec x (sec x + tan x) dx sec x + tan x The substitution u = sec x + tan x gives: Z = du u = ln |u| + c = ln | sec x + tan x| + c Z Exercise 13-8: Evaluate Solution: Z Z 2 tan x dx = tan2 x dx sec2 x − 1 dx = tan x − x + c Z Exercise 13-9: Evaluate tan3 x dx Solution: Z Z 3 tan x dx = Z = tan2 x · tan x dx (sec2 x − 1) · tan x dx Z Z 2 sec x · tan x dx − = Z Z u du − = tan x dx tan x dx where u = tan x = tan2 x − ln | sec x| + c 2 Exercise 13-10: Evaluate the following integrals: Z a) tan4 x dx Z b) Z c) Z d) tan3 x sec2 x dx tan3 x sec x dx tan2 x sec4 x dx Trigonometric Substitutions Using certain trigonometric identities, we can simplify integrals involving square roots: √ a2 − x2 , use x = a sin θ to obtain: π π a| cos θ|, − 6 θ 6 2 2 √ • For a2 + x2 , use x = a tan θ to obtain: π π a| sec θ|, − < θ < 2 2 √ • For x2 − a2 , use x = a sec θ to obtain: π a| tan θ|, 0 6 θ < if x > a 2 π < θ 6 π if x > −a 2 • For Exercise 13-11: Evaluate the following integrals: Z dx √ a) 1 − x2 Z dx √ b) 9 − x2 Z √ c) 4 − x2 dx Z d) Z e) Z f) Z g) √ √ √ x dx 4 − x2 x dx 4 + x2 dx 9 + x2 √ dx x2 − 16 Z h) √ x2 dx 16 − x2 Review Exercises Exercise 13-12: Evaluate the following integrals: Z a) cos6 x dx Z b) cos6 x sin3 x dx Z c) cos(πx) sin(2πx) dx Z d) Z e) Z f) Z g) Z h) Z i) tan8 x sec4 x dx tan3 x sec5 x dx sec3 x dx √ √ x2 dx 25 − x2 dx 9x2 − 16 Z dx √ x 2 x2 − 4 Z dx √ x2 x2 + 1 j) k) cos3 (πx) dx — End of WEEK — Author: Dr. Emre Sermutlu Last Update: December 20, 2016