|\frac120\pi^3 \int\limits_0^\pi \fracx^2 \sin x \cos x\sin^4 x+\cos^4 x dx| is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q53 with Solution

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Correct answer: 15

Let $I=\int_0^{\pi} \frac{x^2\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$ We need to find $\left|\frac{120}{\pi^3}I\right|.$
Use the symmetry substitution $x\mapsto \pi-x$ . Since $\sin(\pi-x)=\sin x,\qquad \cos(\pi-x)=-\cos x,$ we get

=-\frac{(\pi-x)^2\sin x\cos x}{\sin^4 x+\cos^4 x}.$$ Hence $$I=\int_0^{\pi}-\frac{(\pi-x)^2\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ 3. Add the two expressions for $I$: $$2I=\int_0^{\pi}\frac{\bigl(x^2-(\pi-x)^2\bigr)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Now $$x^2-(\pi-x)^2=x^2-(\pi^2-2\pi x+x^2)=2\pi x-\pi^2=\pi(2x-\pi).$$ So $$2I=\pi\int_0^{\pi}\frac{(2x-\pi)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Thus $$I=\frac{\pi}{2}\int_0^{\pi}\frac{(2x-\pi)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ 4. Now observe that $$\frac{d}{dx}(\sin^4 x+\cos^4 x)=4\sin x\cos x(\sin^2 x-\cos^2 x).$$ But a better simplification is $$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x.$$ Also, $$\sin^4 x+\cos^4 x=\frac{3+\cos 4x}{4},$$ though we will use a direct logarithmic derivative form: $$\frac{d}{dx}\ln(\sin^4 x+\cos^4 x) =\frac{4\sin x\cos x(\sin^2 x-\cos^2 x)}{\sin^4 x+\cos^4 x}.$$ This is not directly our integrand, so instead define $$J=\int_0^{\pi}\frac{(2x-\pi)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Let $$u=x-\frac{\pi}{2}.$$ Then $2x-\pi=2u$, and around $x=\frac\pi2$ the denominator is symmetric. A cleaner route is integration by parts using $$2\sin x\cos x=\sin 2x,$$ $$\sin^4 x+\cos^4 x=1-\frac12\sin^2 2x.$$ So $$\frac{\sin x\cos x}{\sin^4 x+\cos^4 x} =\frac{\frac12\sin 2x}{1-\frac12\sin^2 2x}.$$ Now set $$t=\cos 2x,\qquad dt=-2\sin 2x\,dx.$$ Also, $$1-\frac12\sin^2 2x=1-\frac12(1-t^2)=\frac{1+t^2}{2}.$$ Hence $$\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx =\frac{\frac12\sin 2x}{(1+t^2)/2}\,dx =\frac{\sin 2x}{1+t^2}\,dx =-\frac{dt}{2(1+t^2)}.

Therefore an antiderivative is

=-\frac12\tan^{-1}(\cos 2x)+C.$$ 5. Now integrate by parts on $$J=\int_0^{\pi}(2x-\pi)\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Take $$u=2x-\pi,\qquad dv=\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Then $$du=2dx, \qquad v=-\frac12\tan^{-1}(\cos 2x).$$ So $$J=\left.(2x-\pi)\left(-\frac12\tan^{-1}(\cos 2x)\right)\right|_0^{\pi} -\int_0^{\pi}2\left(-\frac12\tan^{-1}(\cos 2x)\right)dx.$$ Thus $$J=\left.-\frac{2x-\pi}{2}\tan^{-1}(\cos 2x)\right|_0^{\pi} +\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx.$$ 6. Evaluate the boundary term. At $x=\pi$: $$2x-\pi=\pi,\quad \cos 2x=\cos 2\pi=1,\quad \tan^{-1}(1)=\frac\pi4.$$ So contribution is $$-\frac{\pi}{2}\cdot \frac\pi4=-\frac{\pi^2}{8}.$$ At $x=0$: $$2x-\pi=-\pi,\quad \cos 0=1,\quad \tan^{-1}(1)=\frac\pi4,$$ so contribution is $$-\frac{-\pi}{2}\cdot \frac\pi4=\frac{\pi^2}{8}.$$ Therefore boundary term equals $$-\frac{\pi^2}{8}-\frac{\pi^2}{8}=-\frac{\pi^2}{4}.$$ Hence $$J=-\frac{\pi^2}{4}+\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx.$$ 7. Now show $$\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx=0.$$ Let $$K=\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx.$$ Using $x\mapsto \pi-x$ gives the same integrand, so instead use $x\mapsto \pi/2-x$ over $[0,\pi/2]$ or directly note periodic symmetry. Split: $$K=\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx+\int_{\pi/2}^{\pi}\tan^{-1}(\cos 2x)dx.$$ In the second integral put $x=\pi-u$: $$\cos 2x=\cos 2u,$$ so this equals another copy over $[0,\pi/2]$, hence $$K=2\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx.$$ Now in this integral put $x\mapsto \frac\pi2-x$: $$\cos\bigl(2(\tfrac\pi2-x)\bigr)=\cos(\pi-2x)=-\cos 2x.$$ Thus $$\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx =\int_0^{\pi/2}\tan^{-1}(-\cos 2x)dx =-\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx,$$ because $\tan^{-1}$ is odd. Therefore the integral is $0$, so $K=0$. Hence $$J=-\frac{\pi^2}{4}.$$ 8. Since $$I=\frac\pi2 J,$$ we get $$I=\frac\pi2\left(-\frac{\pi^2}{4}\right)=-\frac{\pi^3}{8}.$$ 9. Therefore $$\left|\frac{120}{\pi^3}I\right| =\left|\frac{120}{\pi^3}\cdot \left(-\frac{\pi^3}{8}\right)\right| =\left|-15\right|=15.$$ So the required integer is $$\boxed{15}.$$

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View written solutionFree

Correct answer: 15

Let $I=\int_0^{\pi} \frac{x^2\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$ We need to find $\left|\frac{120}{\pi^3}I\right|.$
Use the symmetry substitution $x\mapsto \pi-x$ . Since $\sin(\pi-x)=\sin x,\qquad \cos(\pi-x)=-\cos x,$ we get

=-\frac{(\pi-x)^2\sin x\cos x}{\sin^4 x+\cos^4 x}.$$ Hence $$I=\int_0^{\pi}-\frac{(\pi-x)^2\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ 3. Add the two expressions for $I$: $$2I=\int_0^{\pi}\frac{\bigl(x^2-(\pi-x)^2\bigr)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Now $$x^2-(\pi-x)^2=x^2-(\pi^2-2\pi x+x^2)=2\pi x-\pi^2=\pi(2x-\pi).$$ So $$2I=\pi\int_0^{\pi}\frac{(2x-\pi)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Thus $$I=\frac{\pi}{2}\int_0^{\pi}\frac{(2x-\pi)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ 4. Now observe that $$\frac{d}{dx}(\sin^4 x+\cos^4 x)=4\sin x\cos x(\sin^2 x-\cos^2 x).$$ But a better simplification is $$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x.$$ Also, $$\sin^4 x+\cos^4 x=\frac{3+\cos 4x}{4},$$ though we will use a direct logarithmic derivative form: $$\frac{d}{dx}\ln(\sin^4 x+\cos^4 x) =\frac{4\sin x\cos x(\sin^2 x-\cos^2 x)}{\sin^4 x+\cos^4 x}.$$ This is not directly our integrand, so instead define $$J=\int_0^{\pi}\frac{(2x-\pi)\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Let $$u=x-\frac{\pi}{2}.$$ Then $2x-\pi=2u$, and around $x=\frac\pi2$ the denominator is symmetric. A cleaner route is integration by parts using $$2\sin x\cos x=\sin 2x,$$ $$\sin^4 x+\cos^4 x=1-\frac12\sin^2 2x.$$ So $$\frac{\sin x\cos x}{\sin^4 x+\cos^4 x} =\frac{\frac12\sin 2x}{1-\frac12\sin^2 2x}.$$ Now set $$t=\cos 2x,\qquad dt=-2\sin 2x\,dx.$$ Also, $$1-\frac12\sin^2 2x=1-\frac12(1-t^2)=\frac{1+t^2}{2}.$$ Hence $$\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx =\frac{\frac12\sin 2x}{(1+t^2)/2}\,dx =\frac{\sin 2x}{1+t^2}\,dx =-\frac{dt}{2(1+t^2)}.

Therefore an antiderivative is

=-\frac12\tan^{-1}(\cos 2x)+C.$$ 5. Now integrate by parts on $$J=\int_0^{\pi}(2x-\pi)\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Take $$u=2x-\pi,\qquad dv=\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx.$$ Then $$du=2dx, \qquad v=-\frac12\tan^{-1}(\cos 2x).$$ So $$J=\left.(2x-\pi)\left(-\frac12\tan^{-1}(\cos 2x)\right)\right|_0^{\pi} -\int_0^{\pi}2\left(-\frac12\tan^{-1}(\cos 2x)\right)dx.$$ Thus $$J=\left.-\frac{2x-\pi}{2}\tan^{-1}(\cos 2x)\right|_0^{\pi} +\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx.$$ 6. Evaluate the boundary term. At $x=\pi$: $$2x-\pi=\pi,\quad \cos 2x=\cos 2\pi=1,\quad \tan^{-1}(1)=\frac\pi4.$$ So contribution is $$-\frac{\pi}{2}\cdot \frac\pi4=-\frac{\pi^2}{8}.$$ At $x=0$: $$2x-\pi=-\pi,\quad \cos 0=1,\quad \tan^{-1}(1)=\frac\pi4,$$ so contribution is $$-\frac{-\pi}{2}\cdot \frac\pi4=\frac{\pi^2}{8}.$$ Therefore boundary term equals $$-\frac{\pi^2}{8}-\frac{\pi^2}{8}=-\frac{\pi^2}{4}.$$ Hence $$J=-\frac{\pi^2}{4}+\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx.$$ 7. Now show $$\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx=0.$$ Let $$K=\int_0^{\pi}\tan^{-1}(\cos 2x)\,dx.$$ Using $x\mapsto \pi-x$ gives the same integrand, so instead use $x\mapsto \pi/2-x$ over $[0,\pi/2]$ or directly note periodic symmetry. Split: $$K=\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx+\int_{\pi/2}^{\pi}\tan^{-1}(\cos 2x)dx.$$ In the second integral put $x=\pi-u$: $$\cos 2x=\cos 2u,$$ so this equals another copy over $[0,\pi/2]$, hence $$K=2\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx.$$ Now in this integral put $x\mapsto \frac\pi2-x$: $$\cos\bigl(2(\tfrac\pi2-x)\bigr)=\cos(\pi-2x)=-\cos 2x.$$ Thus $$\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx =\int_0^{\pi/2}\tan^{-1}(-\cos 2x)dx =-\int_0^{\pi/2}\tan^{-1}(\cos 2x)dx,$$ because $\tan^{-1}$ is odd. Therefore the integral is $0$, so $K=0$. Hence $$J=-\frac{\pi^2}{4}.$$ 8. Since $$I=\frac\pi2 J,$$ we get $$I=\frac\pi2\left(-\frac{\pi^2}{4}\right)=-\frac{\pi^3}{8}.$$ 9. Therefore $$\left|\frac{120}{\pi^3}I\right| =\left|\frac{120}{\pi^3}\cdot \left(-\frac{\pi^3}{8}\right)\right| =\left|-15\right|=15.$$ So the required integer is $$\boxed{15}.$$

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