33. (a) Use the reduction formula in Example 6 to show that n-1 sin"x dx /2 = sin"-2x dx n Jo where n ≥ 2 is an integer. (b) Use part (a) to evaluate f/2 sin³x dx and f/2 sin³x dx. (c) Use part (a) to show that, for odd powers of sine, /2 sin2n+1x dx = Jo 2.4.6. 3.5.7. EXAMPLE 6 Prove the reduction formula ...2n' . (2n+1) sin"x dx= n- == cos x sin"-x+ √ sin"-2x dx n n where n > 2 is an integer. SOLUTION Let Then u = sin"-¹x du (n - 1) sin"-2x cos x dx dv = sin x dx v = -cos x so integration by parts gives sin"-2x cos²x dx sin"x dx = cos x sin"-¹x + (n - - 1) √ sin"- = Since cos²x 1 - sin²x, we have - | sin"xdx = −cos x sin" ! + (n − 1) | sin"-x dx – (n − 1) sinx dx As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side. Thus we have nfsin"x dx = cos x sin"-'x + (n - 1) sin"-2x dx or sin"x dx n 1 cos x sin"-x+ √ sin"-2x dx n n The reduction formula 7 is useful because by using it repeatedly we could even- tually express sin"x dx in terms of sin x dx (if n is odd) or f (sin x)° dx = √ dx (if n is even).