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Trigonometric Integrals R The book gives :::: lots of rules/strategies for Trigonometric Integrals (e.g. tan2 x cos x dx). You are welcome to read them but do NOT memorize. Use logic instead, as done below! ::::: • Recall the following natural pairings (in terms of derivatives): sin ↔ cos tan ↔ sec cot ↔ csc If the integrand does not contain natural pairings, then rewrite it so that it does. 2x sin2 x sin2 x or tan2 x cos x = tan2 x sec1 x = tan . Eg: tan2 x cos x = cos 2 x cos x = cos x sec x • Next try a u-du substitution. There will be two naturals choice for u. ◦ If the integrand involves sine and/or cosine, then try u = cos x or u = sin x. ◦ If the integrand involves tangent and/or secant, then try u = tan x or u = sec x. ◦ If the integrand involves cotangent and/or cosecant, then try u = cot x or u = csc x. • First isolate off your du (box him and don’t touch him anymore) and ::::: then try to express what’s left in terms of only u. ::::: The following will be helpful (and you must memorize them): 1 − cos(2x) 1 + cos(2x) and sin2 x = (half-angle ( 12 ∠) formulas) cos2 x = 2 2 cos(2x) = cos2 x − sin2 x and sin(2x) = 2 sin x cos x (double-angle (2∠) formulas) cos2 x sin2 x 1 + = =⇒ 1 + tan2 x = sec2 x 2 2 cos x cos x cos2 x cos2 x sin2 x 1 cos2 x + sin2 x = 1 =⇒ + = =⇒ cot2 x + 1 = csc2 x 2 2 2 sin x sin x sin x R R R I. For integrands of the form: sin (αx) cos (βx) dx or sin (αx) sin (βx) dx or cos (αx) cos (βx) dx where α, β ∈ R and α:::::: 6= β (see Example 5) use the trig identities (these 3 you do not have to memorize) (1) sin (A) cos (B) = 21 [sin(A − B) + sin(A + B)] cos2 x + sin2 x = 1 =⇒ (2) sin (A) sin (B) = (3) cos (A) cos (B) = 1 [cos(A 2 1 [cos(A 2 − B) − cos(A + B)] − B) + cos(A + B)] . u=cos x Example 4. sin4 x dx. u-du sub does not work (why? e.g.: sin4 x dx = − sin3 x [− sin x dx]). R For sinn x cosm x dx, with both m, n ∈ {0, 2, 4, 6, . . .}, use the half-angle formulas. ::::: 2 Z Z Z Z ( 12 ∠) A A 1 1 − cos(2x) 4 2 2 dx = [1 − 2 cos(2x) + cos2 (2x)] dx sin x dx = [sin x] dx = 2 4 Z ( 21 ∠) 1 1 + cos(4x) = 1 − 2 cos(2x) + dx 4 2 Z Z Z Z 1 1 1 1 1 1 = dx − 2 cos(2x) dx + · dx + · cos(4x) dx 4 4 4 2 4 2 Z Z Z 1 1 1 1 1 1 · = 1+ dx − · cos(2x) [2dx] + · cos(4x) [4dx] 4 2 4 4 2 4 1 1 3 = x − sin(2x) + sin(4x) + C 8 4 32 R R R to here is fine now but in trig-sub we’ll want all angles to be (1x) so let’s do it . . . note: sin (4x) = sin (2 (2x)) 3 1 1 x − (2 sin x cos x) + [(2) sin(2x) cos(2x)] + C 8 4 32 1 1 (2∠) 3 = x − (sin x · cos x) + (2) (2 sin x · cos x) cos2 x − sin2 x + C 8 2 32 Good enough for now. R Example 5. sin (2x) cos (3x) dx. Hint: use (1) above with A = 2x, B = 3x. Soln is on bottom of page 4. (2∠) = 17.01.02 (yr.mn.dy) Page 1 of 4 Techniques of Integration Trigonometric Integrals Z sin7 x dx Example 1. u = cos x du = − sin x dx try 1: First box off your du: ::::: u = cos x Z u = sin x du = cos x dx or :: so du = − sin x dx Z 7 6 sin x dx = − sin x − sin x dx . Then try to express what’s left, i.e. sin6 x , in terms of only u, i.e., in terms of only cos x: ::::: sin6 x = (sin x)6 = sin2 x 3 = 1 − cos2 x 3 . So: Z Z 7 sin6 x − sin x dx Z h 3 i =− sin2 x − sin x dx Z h 3 i 2 =− − sin x dx 1 − cos x sin x dx = − , did it. (∗Ex.1 for u = cos x) in Examples 1–3, once you get the the ∗-ed :::::: clever::::: way ::: to ::::: look ::: at :::: the :::::::: integral::: so:::: can:::: do ::::: u-du:::: sub, it’s smooth sailing from here on out! Z h 3 i =− 1 − u2 du Z A = u6 − 3u4 + 3u2 − 1 du u7 3u5 3u3 − + −u+C 7 5 3 cos7 x 3 cos5 x = − + cos3 x − cos x + C 7 5 = try 2: u = sin x so du = cos x dx First box off your du: ::::: Z 7 sin x dx = Z [???] would want here cos x dx . The substitution with u = sin x doesn’t work because there is no cos x in the integrand to help make the needed du. 17.01.02 (yr.mn.dy) /////////// Page 2 of 4 Techniques of Integration Trigonometric Integrals Z Example 2. sin4 (17x) cos3 (17x) dx u = cos(17x) du = −17 sin(17x) dx try 1: u = sin(17x) du = 17 cos(17x) dx or :: u = cos (17x) so du = −17 sin (17x) dx First box off your du: ::::: Z Z 3 −1 4 3 sin (17x) cos (17x) dx = cos (17x) sin3 (17x) −17 sin(17x) dx . 17 3 Then try to express what’s left, i.e. cos (17x) sin3 (17x) , in terms of only u, i.e., in terms of only ::::: cos (17x) : cos3 (17x) sin3 (17x) = cos3 (17x) sin2 (17x) sin(17x) = cos3 (17x) 1 − cos2 (17x) sin(17x) . Cannot do, there is an “extra sin” try 2: /////////// u = sin (17x) so du = 17 cos (17x) dx First box off your du: ::::: Z Z 4 1 4 3 sin (17x) cos (17x) dx = sin (17x) cos2 (17x) 17 cos(17x) dx . 17 4 Then try to express what’s left, i.e. sin (17x) cos2 (17x) , in terms of only u, i.e., in terms of only ::::: sin(17x): sin4 (17x) cos2 (17x) = sin4 (17x) 1 − sin2 (17x) So: Z , did it. sin4 (17x) cos3 (17x) dx Z 4 1 sin (17x) cos2 (17x) 17 cos(17x) dx = 17 Z 4 1 = sin (17x) 1 − sin2 (17x) 17 cos(17x) dx (∗Ex.2 for u = sin(17x)) 17 Z 4 1 = u 1 − u2 du 17 Z A 1 = (u4 − u6 ) du 17 1 u5 u7 1 sin5 (17x) sin7 (17x) = − +C = − +C . 17 5 7 17 5 7 17.01.02 (yr.mn.dy) Page 3 of 4 Techniques of Integration Trigonometric Integrals Z Example 3. tan θ sec4 θ dθ try 1: u = tan θ du = sec2 θ dθ so: u = tan θ so or :: u = sec θ du = sec θ tan θ dθ du = sec2 θ dθ First box off your du: ::::: Z Z 4 tan θ sec θ dθ = tan θ sec2 θ sec2 θ dθ . Then try to express what’s left, i.e. [tan θ sec2 θ], in terms of only u, i.e. in terms of only tan θ: ::::: tan θ sec2 θ = tan θ 1 + tan2 θ did it. So: Z Z 4 tan θ sec θ dθ = , tan θ sec2 θ sec2 θ dθ Z tan θ 1 + tan2 θ sec2 θ dθ (∗Ex.3 for u = tan θ) Z Z A u2 u4 tan2 θ tan4 θ 2 (u + u3 ) du = + +C = + +C = u(1 + u ) du = 2 4 2 4 = try 2: u = sec θ so du = sec θ tan θ dθ First box off your du: ::::: Z 4 Z tan θ sec θ dθ = 3 sec θ sec θ tan θ dθ . , Then try to express what’s left, i.e. [sec3 θ], in terms of only u, i.e. in terms of only sec θ. easy ::::: So: Z Z 3 4 tan θ sec θ dθ = (∗Ex.3 for u = sec θ) sec θ sec θ tan θ dθ Z 3 u4 sec4 θ = u du = +K = +K 4 4 Our answers from try 1 and try 2 should be the same . . . and they are since: 2 2 A 1 T 1 sec4 θ +K = sec2 θ + K = 1 + tan2 θ + K 4 4 4 2 A tan θ A 1 tan4 θ 1 2 4 1 + 2 tan θ + tan θ + K = + + K+ . = 4 2 4 4 So 1 C=K+ . 4 Now onto Example 4 (on page 1) and below is the Solution to Example 5. Using trig identities: first (1) sin (A) cos (B) = 21 [sin(A − B) + sin(A + B)] and then last cos(−x) = cos x. Z Z Z 1 (1) 1 sin (2x) cos (3x) dx = [sin (2x − 3x) + sin (2x + 3x)] dx = [sin (−x) + sin (5x)] dx 2 2 Z Z 1 1 1 = (−1) sin (−1x) (−1 dx) + sin (5x) (5 dx) 2 2 5 cos (−x) cos(5x) cos x cos(5x) − +C = − +C . = 2 10 2 10 17.01.02 (yr.mn.dy) Page 4 of 4 Techniques of Integration