If f is continuous on [0, π], use the substitution u = π-x to show that π [" xf(sin(x)) dx = xf(sin(x)) dx = So we have, 元 [xf(si Let u = π - x. Then du = ---Select--- . When x = π, u = 0, and when x = 0, u = ---Select--- ✓ Use U-substitution to replace the values in the integral. [² - [° (π - L) r (sin( [ u) π 7. * F (sir π = ["² (72 - u ) ₁( sin ( (π- Recall the identity that sin(π - u) = ["xf(sin(x)) dx = [(x - u)r (sin( [ = π 2 6.² xf(sin(x)) dx = π *S*H f(sin(x)) dx. = 7 *. [**f(sin(u)) du - - π ² [²x 2 x f(sin(x)) dx = π 10 *** (sin([ f(sin(x)) dx - sin(u) and make the appropriate substitution and continue the proof. ])) du [*ur(sin( [ - [*x r( sin ( [ f(sin(x)) dx - *6*x r( sin ( [ (- du) dx du ])) du 1)) dx ])) dx → →>>

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