Practicing Success
If $\int\limits_0^\pi x f(\sin x) d x=A \int\limits_0^{\pi / 2} f(\sin x) d x$, then $A$ |
$2 \pi$ $\pi$ $\pi / 4$ 0 |
$\pi$ |
Let $I=\int\limits_0^\pi x f(\sin x) d x$ $I=\int\limits_0^\pi(\pi-x) f(\sin (\pi-x)) d x$ $\left[∵ \int\limits_0^a f(x) d x=\int\limits_0^a f(a-x) d x\right]$ $\Rightarrow I=\pi \int\limits_0^\pi f(\sin x) d x-I$ $\Rightarrow I=\frac{\pi}{2} \int\limits_0^\pi f(\sin x) d x$ $\Rightarrow I=\frac{\pi}{2} \times 2 \int\limits_0^{\pi / 2} f(\sin x) d x$ $\left[\begin{array}{cc}∵ \int\limits_0^{2 a} f(x) d x=2 \int\limits_0^a f(x) d x \text { if, } f(2 a-x)=f(x)\end{array}\right]$ $\Rightarrow I=\pi \int\limits_0^{\pi / 2} f(\sin x)$ ∴ $\int\limits_0^\pi x f(\sin x) d x=A \int\limits_0^{\pi / 2} f(\sin x) d x$ $\Rightarrow \pi \int\limits_0^{\pi / 2} f(\sin x) d x=A \int\limits_0^{\pi / 2} f(\sin x) d x$ $\Rightarrow A=\pi$ |