Integrating Powers of Sine and Cosine

Integrating Powers of Sine and Cosine There are three really important identities that you really ought to commit to memory, since so many other identities stem from these: • sin2 (θ) + cos2 (θ) = 1 • sin(2θ) = 2 sin(θ) cos(θ) • cos(2θ) = cos2 (θ) − sin2 (θ) Using the third identity, we get the half angle formulas: cos(2θ) = cos2 (θ) − (1 − cos2 (θ)) = 2 cos2 (θ) − 1 ⇔ cos2 (θ) = 1 + cos(2θ) 2 cos(2θ) = (1 − sin2 (θ)) − sin2 (θ) = 1 − 2 sin2 (θ) ⇔ sin2 (θ) = 1 − cos(2θ) 2 You should know these formulas. Integrate Z sinm (x) cosn (x) dx when m or n is odd (or both) The idea: Take the term with the odd power, and reserve one to go with dx in preparation for u, du substitution. Here are a couple of examples: • Z sinm (x) cos(x) dx (In this case, m can be even or odd, and n = 1) SOLUTION: Let u = sin(x) so that du = cos(x) dx, and the integral becomes (for m > 0) Z 1 um du = sinm+1 (x) + C m+1 • Z sin3 (x) cos2 (x) dx SOLUTION: Reserve one sin(x) for the substitution, then write everything else in terms of cos(x). We will let u = cos(x) so du = − sin(x) dx: Z 2 2 sin (x) cos (x) (sin(x) dx) = − • Similarly: Z Z (1 − u2 )u2 du = sin3 (x) dx = Z Z (1 − cos2 (x)) cos2 (x) (sin(x) dx) = 1 1 cos5 (x) − cos3 (x) + C 5 3 sin2 (x)[sin(x) dx] = 1 Z (1 − cos2 (x))[sin(x) dx] Integrating when both powers are even Z In this case, we integrate sinm (x) cosn (x) dx, when both m and n are even. The idea: Use the half angle formulas to reduce the exponents, hopefully to something we can integrate. Often, this step is taken multiple times. • • Z Z 2 sin (x) dx = Z 1 1 1 (1 − cos(2x)) dx = x − sin(2x) + C 2 2 4 sin2 (x) cos2 (x) dx Method 1: Use half angle formulas: sin2 (x) cos2 (x) = Similarly, cos2 (2x) = 1+cos(4x) , 2 sin2 (x) cos2 (x) = 1 − cos(2x) 1 + cos(2x) 1 = (1 − cos2 (2x)) 2 2 4 so this further reduces to: 1 1 1 + cos(4x) 1 1 1 1 1 − · = − − cos(4x) = − cos(4x) 4 4 2 4 8 8 8 8 And this is easy to integrate: 18 x − 1 32 sin(4x) + C Method 2: Use the identity for sin(2x): 2 2 2 sin (x) cos (x) = (sin(x) cos(x)) = sin(2x) 2 Exercises 1. 2. 3. 4. 5. 6. Z sin2 (x) cos3 (x) dx Z sin7 (x) cos5 (x) dx Z cos2 (x) dx Z sin4 (x) dx Z x sin2 (x) dx Z x sin3 (x) dx 2 !2 1 1 = sin2 (2x) = 4 4 1 − cos(4x) 2 ! Solutions 1. Z sin2 (x) cos3 (x) dx SOLUTION: Using the odd power, we reserve cos(x) with dx to use as du. Rewrite what remains in terms of sines (using the Pythagorean Identity): Z 2 2 sin (x) cos (x) [cos(x) dx] = Z sin2 (x)(1 − sin2 (x)) [cos(x) dx] Now, take u = sin(x) so du = cos(x) dx, and substitute: Z 2. Z 1 1 1 1 u2 − u4 du = u3 − u5 = sin3 (θ) − sin5 (θ) + C 3 5 3 5 sin7 (x) cos5 (x) dx SOLUTION: We can choose to reserve either a sine or a cosine, since both powers are odd. Choosing a cos(x), we have: Z 7 4 sin (x) cos (x) [cos(x) dx] = Z sin7 (x)(1 − sin2 (x))2 [cos(x) dx] Make the substitution with u = sin(x) and we get (expand the square): Z u7 (1 − 2u2 + u4 ) du = Z u7 − 2u9 + u11 du which is simple to integrate: 1 1 1 8 sin (x) − sin10 (x) + sin12 (x) + C 8 5 12 3. Z cos2 (x) dx SOLUTION: For an even power use the half angle formula: Z 4. Z 1 + cos(2x) 1 1 dx = x + sin(2x) + C 2 2 4 sin4 (x) dx SOLUTION: Use the half angle formula (twice): Z 1 − cos(2x) 2 !2 dx = Z 1 (1 − 2 cos(2x) + cos2 (2x)) dx 4 1 + cos(4x) , and we have: 2 Z 3 1 1 3 1 1 − cos(2x) + cos(4x) dx = x − sin(2x) + sin(4x) + C 8 2 8 8 4 32 Use the half angle formula again for cos2 (2x) = 3 5. Z x sin2 (x) dx There are two issues here: Because of the leading x, we’ll need to use “integration by parts”, and because of the sin2 (x), we’ll need to use the half angle formula. For the standard integration by parts, we need to assign values to u and dv, then find du and v: u=x du = 1 dx Z 1 1 1Z 2 2 1 − cos(2x) dx = x − sin(2x) dv = sin (x) dx ⇒ v = sin (x) dx = 2 2 4 6. Z x sin3 (x) dx 4