\int\limits_0^\pi xf( \sin x )dx is equal to

\pi \int\limits_0^\pi /2 f( \cos x )dx

\pi \int\limits_0^\pi f( \cos x )dx

\,\pi \int\limits_0^\pi f( sinx )dx

\pi \over 2\int\limits_0^\pi /2 f( sinx )dx

Definite Integration PYQ — JEE Main 2006 · Shift 0 · Q61 with Solution

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Correct answer: D

Let $I=\int_0^{\pi} x\,f(\sin x)\,dx.$
Use the standard property of definite integrals: If $I=\int_0^a g(x)\,dx,$ then by substituting $x\mapsto a-x$ , $I=\int_0^a g(a-x)\,dx.$

So here, $I=\int_0^{\pi} (\pi-x) f(\sin(\pi-x))\,dx.$ Since $\sin(\pi-x)=\sin x,$ we get $I=\int_0^{\pi} (\pi-x) f(\sin x)\,dx.$

Add the two expressions for $I$ : $I=\int_0^{\pi} x f(\sin x)\,dx,$ and $I=\int_0^{\pi} (\pi-x) f(\sin x)\,dx.$

Thus,

=\int_0^{\pi} \pi f(\sin x)\,dx.$$ So, $$2I=\pi\int_0^{\pi} f(\sin x)\,dx,$$ which gives $$I=\frac{\pi}{2}\int_0^{\pi} f(\sin x)\,dx.$$ 4. Now use symmetry again on $$J=\int_0^{\pi} f(\sin x)\,dx.$$ Because $\sin(\pi-x)=\sin x$, $$J=2\int_0^{\pi/2} f(\sin x)\,dx.$$ Hence, $$I=\frac{\pi}{2}\cdot 2\int_0^{\pi/2} f(\sin x)\,dx =\pi\int_0^{\pi/2} f(\sin x)\,dx.$$ 5. Also, on $[0,\pi/2]$, $$\sin x=\cos\left(\frac{\pi}{2}-x\right),$$ so $$\int_0^{\pi/2} f(\sin x)\,dx=\int_0^{\pi/2} f(\cos x)\,dx.$$ Therefore, $$I=\pi\int_0^{\pi/2} f(\cos x)\,dx.$$ 6. Compare with options: - A: $$\pi\int_0^{\pi} f(\cos x)\,dx$$ — not equal. - B: $$\pi\int_0^{\pi} f(\sin x)\,dx$$ — missing factor $\tfrac12$. - C: $$\frac{\pi}{2}\int_0^{\pi/2} f(\sin x)\,dx$$ — too small by factor $2$. - D: $$\pi\int_0^{\pi/2} f(\cos x)\,dx$$ — correct. Hence the correct option is $$\boxed{D}.$$

View written solutionFree

Correct answer: D

Let $I=\int_0^{\pi} x\,f(\sin x)\,dx.$
Use the standard property of definite integrals: If $I=\int_0^a g(x)\,dx,$ then by substituting $x\mapsto a-x$ , $I=\int_0^a g(a-x)\,dx.$

So here, $I=\int_0^{\pi} (\pi-x) f(\sin(\pi-x))\,dx.$ Since $\sin(\pi-x)=\sin x,$ we get $I=\int_0^{\pi} (\pi-x) f(\sin x)\,dx.$

Add the two expressions for $I$ : $I=\int_0^{\pi} x f(\sin x)\,dx,$ and $I=\int_0^{\pi} (\pi-x) f(\sin x)\,dx.$

Thus,

=\int_0^{\pi} \pi f(\sin x)\,dx.$$ So, $$2I=\pi\int_0^{\pi} f(\sin x)\,dx,$$ which gives $$I=\frac{\pi}{2}\int_0^{\pi} f(\sin x)\,dx.$$ 4. Now use symmetry again on $$J=\int_0^{\pi} f(\sin x)\,dx.$$ Because $\sin(\pi-x)=\sin x$, $$J=2\int_0^{\pi/2} f(\sin x)\,dx.$$ Hence, $$I=\frac{\pi}{2}\cdot 2\int_0^{\pi/2} f(\sin x)\,dx =\pi\int_0^{\pi/2} f(\sin x)\,dx.$$ 5. Also, on $[0,\pi/2]$, $$\sin x=\cos\left(\frac{\pi}{2}-x\right),$$ so $$\int_0^{\pi/2} f(\sin x)\,dx=\int_0^{\pi/2} f(\cos x)\,dx.$$ Therefore, $$I=\pi\int_0^{\pi/2} f(\cos x)\,dx.$$ 6. Compare with options: - A: $$\pi\int_0^{\pi} f(\cos x)\,dx$$ — not equal. - B: $$\pi\int_0^{\pi} f(\sin x)\,dx$$ — missing factor $\tfrac12$. - C: $$\frac{\pi}{2}\int_0^{\pi/2} f(\sin x)\,dx$$ — too small by factor $2$. - D: $$\pi\int_0^{\pi/2} f(\cos x)\,dx$$ — correct. Hence the correct option is $$\boxed{D}.$$