Question #17bd1

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Answer 1 · 2016-07-08T08:31:51

See.the explanation, for proof.

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

and 0, and dx = $- d t$ $-dt$ . .

Now, $I = - \int (π - t) f (sin (π - t)) d t$ $I=-int(pi-t)f(sin(pi-t)) dt$ , from $t = π$ $t= pi$ to t = 0.

$= π \int f (sin t) d t - \int t f (sin t) d t$ $=pi int f(sin t) dt-int t f(sin t) dt$ , from t = 0 to $π$ $pi$ ,

using $sin (π - t) = sin t$ $sin (pi-t)=sin t$ ,

$= π \int f (sin x) d x$ $=pi int f(sin x) dx$ , from x = o to x = $π - I$ $pi -I$ .

Thus, $I = \frac{π}{2} \int f (sin x) d x$ $I = pi/2 int f(sin x) dx$ , from x = 0 to x = $π$ $pi$ .