Question #17bd1

Jul 8, 2016

See.the explanation, for proof.

Explanation:

Let I be the definite integral and $x = \pi - t$. The limits become $\pi$

and 0, and dx = $- \mathrm{dt}$. .

Now, $I = - \int \left(\pi - t\right) f \left(\sin \left(\pi - t\right)\right) \mathrm{dt}$, from $t = \pi$ to t = 0.

$= \pi \int f \left(\sin t\right) \mathrm{dt} - \int t f \left(\sin t\right) \mathrm{dt}$, from t = 0 to $\pi$,

using $\sin \left(\pi - t\right) = \sin t$,

$= \pi \int f \left(\sin x\right) \mathrm{dx}$, from x = o to x =$\pi - I$.

Thus,$I = \frac{\pi}{2} \int f \left(\sin x\right) \mathrm{dx}$, from x = 0 to x =$\pi$.