Practicing Success
$\int\limits_{0}^{\pi}f(sin x) dx$ is equal to |
$\pi\int\limits_{0}^{\pi}f(sinx)dx$ $\pi\int\limits_{0}^{\frac{\pi}{2}}f(sinx)dx$ $2\pi\int\limits_{0}^{\frac{\pi}{2}}f(sinx)dx$ none of these |
$\pi\int\limits_{0}^{\frac{\pi}{2}}f(sinx)dx$ |
$I=\int\limits_{0}^{\pi}f(sin x) dx$ $=\int\limits_{0}^{\pi}(\pi-x)f\,sin(\pi-x)dx=\pi\int\limits_{0}^{\pi}f(sinx)dx-I$ $⇒I=\frac{\pi}{2}f(sinx)dx$. Again, $I =\frac{\pi}{2}\int\limits_{0}^{\pi}f(sinx) dx = 2\frac{\pi}{2}\int\limits_{0}^{\pi/2}f(sinx) dx = \pi\int\limits_{0}^{\pi/2}f(sinx) dx$. Hence (B) is the correct answer. |