JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Definite Integrals

Let $I={\int }_0^{\pi }\frac{x^2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$ and $g(x)=\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}$, which has period $\pi$ and satisfies $g(\pi -x)=-g(x)$.

Applying $x\to \pi -x$ to $I$:

$I={\int }_0^{\pi }(\pi -x{)}^{2}g(\pi -x)dx=-{\int }_0^{\pi }({\pi }^2-2\pi x+x^2)g(x)dx=-{\pi }^2{\int }_0^{\pi }gdx+2\pi {\int }_0^{\pi }xgdx-I$.

Since $g(\pi -x)=-g(x)$, ${\int }_0^{\pi }gdx=0$ (antisymmetric about the midpoint). Writing $J={\int }_0^{\pi }xg(x)dx$:

$I=2\pi J-I\Rightarrow I=\pi J$.

To evaluate $J$, substitute $x=\frac{\pi }{2}+t$ ($t$ from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$); one finds $g\left(\frac{\pi }{2}+t\right)=-g(t)$ where $g(t)=\frac{\sin t\cos t}{{\sin }^{4}t+{\cos }^{4}t}$ is an odd function of $t$. This gives

$J=-2{\int }_0^{\pi /2}tg(t)dt=-2K$, $K={\int }_0^{\pi /2}tg(t)dt$.

Applying $t\to \frac{\pi }{2}-t$ inside $K$ (here $g\left(\frac{\pi }{2}-t\right)=g(t)$) gives $K=\frac{\pi }{2}{\int }_0^{\pi /2}g(t)dt-K$, so $K=\frac{\pi }{4}L$, where $L={\int }_0^{\pi /2}\frac{\sin t\cos t}{{\sin }^{4}t+{\cos }^{4}t}dt$.

Substitute $u={\sin }^{2}t$, $du=2\sin t\cos tdt$; then ${\sin }^{4}t+{\cos }^{4}t=u^2+(1-u{)}^2=2u^2-2u+1$:

$L=\frac{1}{2}{\int }_0^1\frac{du}{2u^2-2u+1}$. Completing the square, $2u^2-2u+1=2{\left(u-\frac{1}{2}\right)}^2+\frac{1}{2}$; with $v=2u-1$,

${\int }_0^1\frac{du}{2u^2-2u+1}={\int }_{-1}^1\frac{dv}{v^2+1}=\frac{\pi }{4}-\left(-\frac{\pi }{4}\right)=\frac{\pi }{2}$, so $L=\frac{\pi }{4}$.

Then $K=\frac{\pi }{4}\cdot \frac{\pi }{4}=\frac{{\pi }^2}{16}$, $J=-2K=-\frac{{\pi }^2}{8}$, and $I=\pi J=-\frac{{\pi }^3}{8}$.

$\left|\frac{120}{{\pi }^3}\cdot I\right|=\left|\frac{120}{{\pi }^3}\cdot \left(-\frac{{\pi }^3}{8}\right)\right|=\frac{120}{8}=15$.