If \int\limits_0^\pi xf( \sin x )dx = A\int\limits_0^\pi /2 f( \sin x )dx, then A is

Definite Integration PYQ — JEE Main 2004 · Shift 0 · Q86 with Solution

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Correct answer: B

Let $I=\int_0^\pi x\,f(\sin x)\,dx.$ We need to express this in the form $I=A\int_0^{\pi/2} f(\sin x)\,dx.$
Use the standard substitution $x\mapsto \pi-x$ in the same integral. Since $\sin(\pi-x)=\sin x,$ we get $I=\int_0^\pi (\pi-x)f(\sin x)\,dx.$
Add the two expressions for $I$ :

=\int_0^\pi \pi f(\sin x)\,dx.$$ So, $$I=\frac{\pi}{2}\int_0^\pi f(\sin x)\,dx.$$ 4. Now use symmetry again for the integral without $x$: $$\int_0^\pi f(\sin x)\,dx=2\int_0^{\pi/2} f(\sin x)\,dx,$$ because $\sin x$ is symmetric about $\pi/2$. Thus, $$I=\frac{\pi}{2}\cdot 2\int_0^{\pi/2} f(\sin x)\,dx =\pi\int_0^{\pi/2} f(\sin x)\,dx.$$ 5. Comparing with $$\int_0^\pi x f(\sin x)dx=A\int_0^{\pi/2} f(\sin x)dx,$$ we get $$A=\pi.$$ 6. Option check: - A: $2\pi$ — incorrect - B: $\pi$ — correct - C: $\pi/4$ — incorrect - D: $0$ — incorrect

View written solutionFree

Correct answer: B

Let $I=\int_0^\pi x\,f(\sin x)\,dx.$ We need to express this in the form $I=A\int_0^{\pi/2} f(\sin x)\,dx.$
Use the standard substitution $x\mapsto \pi-x$ in the same integral. Since $\sin(\pi-x)=\sin x,$ we get $I=\int_0^\pi (\pi-x)f(\sin x)\,dx.$
Add the two expressions for $I$ :

=\int_0^\pi \pi f(\sin x)\,dx.$$ So, $$I=\frac{\pi}{2}\int_0^\pi f(\sin x)\,dx.$$ 4. Now use symmetry again for the integral without $x$: $$\int_0^\pi f(\sin x)\,dx=2\int_0^{\pi/2} f(\sin x)\,dx,$$ because $\sin x$ is symmetric about $\pi/2$. Thus, $$I=\frac{\pi}{2}\cdot 2\int_0^{\pi/2} f(\sin x)\,dx =\pi\int_0^{\pi/2} f(\sin x)\,dx.$$ 5. Comparing with $$\int_0^\pi x f(\sin x)dx=A\int_0^{\pi/2} f(\sin x)dx,$$ we get $$A=\pi.$$ 6. Option check: - A: $2\pi$ — incorrect - B: $\pi$ — correct - C: $\pi/4$ — incorrect - D: $0$ — incorrect