If \int_-\frac\pi2^\frac\pi2 \frac96 x^2 \cos ^2 x(1+e^x) d x=\pi(\alpha \pi^2+\beta), \alpha, \beta \in \mathbbZ , then (\alpha+\beta)^2 equals

Definite Integration PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q32 with Solution

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Correct answer: B

Let $I=\int_{-\pi/2}^{\pi/2} \frac{96x^2\cos^2 x}{1+e^x}\,dx.$

We use the standard symmetry trick: if $I=\int_{-a}^{a}\frac{f(x)}{1+e^x}\,dx,$ then on substituting $x\mapsto -x$ ,

=\int_{-a}^{a}\frac{e^x f(-x)}{1+e^x}\,dx.$$ Adding both forms gives $$2I=\int_{-a}^{a}\frac{f(x)+e^x f(-x)}{1+e^x}\,dx.$$ When $f$ is even, $f(-x)=f(x)$, so $$2I=\int_{-a}^{a}f(x)\,dx \quad\Rightarrow\quad I=\frac12\int_{-a}^{a}f(x)\,dx.$$ 2. Here $$f(x)=96x^2\cos^2 x,$$ which is even since $x^2$ and $\cos^2 x$ are both even. Hence $$I=\frac12\int_{-\pi/2}^{\pi/2}96x^2\cos^2 x\,dx =48\int_{-\pi/2}^{\pi/2}x^2\cos^2 x\,dx.$$ Again the integrand is even, so $$I=96\int_0^{\pi/2}x^2\cos^2 x\,dx.$$ 3. Use $$\cos^2 x=\frac{1+\cos 2x}{2}.$$ Then $$I=96\int_0^{\pi/2}x^2\cdot \frac{1+\cos 2x}{2}\,dx =48\int_0^{\pi/2}x^2\,dx+48\int_0^{\pi/2}x^2\cos 2x\,dx.$$ So we compute the two parts separately. 4. First part: $$\int_0^{\pi/2}x^2\,dx=\left[\frac{x^3}{3}\right]_0^{\pi/2}=\frac{\pi^3}{24}.$$ Therefore $$48\int_0^{\pi/2}x^2\,dx=48\cdot \frac{\pi^3}{24}=2\pi^3.$$ 5. Second part: Let $$J=\int_0^{\pi/2}x^2\cos 2x\,dx.$$ Integrate by parts: Take $u=x^2$, $dv=\cos 2x\,dx$. Then $$du=2x\,dx,\qquad v=\frac12\sin 2x.$$ So $$J=\left[\frac{x^2}{2}\sin 2x\right]_0^{\pi/2}-\int_0^{\pi/2}x\sin 2x\,dx.$$ Boundary term is $0$, since $\sin 0=\sin \pi=0$. Hence $$J=-\int_0^{\pi/2}x\sin 2x\,dx.$$ Again integrate by parts: Take $u=x$, $dv=\sin 2x\,dx$. Then $$du=dx,\qquad v=-\frac12\cos 2x.$$ Thus $$\int_0^{\pi/2}x\sin 2x\,dx=\left[-\frac{x}{2}\cos 2x\right]_0^{\pi/2}+\frac12\int_0^{\pi/2}\cos 2x\,dx.$$ Now $$\left[-\frac{x}{2}\cos 2x\right]_0^{\pi/2}=-\frac{\pi/2}{2}\cos \pi-0=\frac{\pi}{4},$$ and $$\frac12\int_0^{\pi/2}\cos 2x\,dx=\frac12\left[\frac12\sin 2x\right]_0^{\pi/2}=0.$$ So $$\int_0^{\pi/2}x\sin 2x\,dx=\frac{\pi}{4},$$ therefore $$J=-\frac{\pi}{4}.$$ Hence $$48J=48\left(-\frac{\pi}{4}\right)=-12\pi.$$ 6. Therefore $$I=2\pi^3-12\pi=\pi(2\pi^2-12).$$ Comparing with $$I=\pi(\alpha\pi^2+\beta),$$ we get $$\alpha=2,\qquad \beta=-12.$$ So $$(\alpha+\beta)^2=(2-12)^2=(-10)^2=100.$$ 7. Hence the correct option is $$\boxed{\text{B: }100}.$$

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View written solutionFree

Correct answer: B

Let $I=\int_{-\pi/2}^{\pi/2} \frac{96x^2\cos^2 x}{1+e^x}\,dx.$

We use the standard symmetry trick: if $I=\int_{-a}^{a}\frac{f(x)}{1+e^x}\,dx,$ then on substituting $x\mapsto -x$ ,

=\int_{-a}^{a}\frac{e^x f(-x)}{1+e^x}\,dx.$$ Adding both forms gives $$2I=\int_{-a}^{a}\frac{f(x)+e^x f(-x)}{1+e^x}\,dx.$$ When $f$ is even, $f(-x)=f(x)$, so $$2I=\int_{-a}^{a}f(x)\,dx \quad\Rightarrow\quad I=\frac12\int_{-a}^{a}f(x)\,dx.$$ 2. Here $$f(x)=96x^2\cos^2 x,$$ which is even since $x^2$ and $\cos^2 x$ are both even. Hence $$I=\frac12\int_{-\pi/2}^{\pi/2}96x^2\cos^2 x\,dx =48\int_{-\pi/2}^{\pi/2}x^2\cos^2 x\,dx.$$ Again the integrand is even, so $$I=96\int_0^{\pi/2}x^2\cos^2 x\,dx.$$ 3. Use $$\cos^2 x=\frac{1+\cos 2x}{2}.$$ Then $$I=96\int_0^{\pi/2}x^2\cdot \frac{1+\cos 2x}{2}\,dx =48\int_0^{\pi/2}x^2\,dx+48\int_0^{\pi/2}x^2\cos 2x\,dx.$$ So we compute the two parts separately. 4. First part: $$\int_0^{\pi/2}x^2\,dx=\left[\frac{x^3}{3}\right]_0^{\pi/2}=\frac{\pi^3}{24}.$$ Therefore $$48\int_0^{\pi/2}x^2\,dx=48\cdot \frac{\pi^3}{24}=2\pi^3.$$ 5. Second part: Let $$J=\int_0^{\pi/2}x^2\cos 2x\,dx.$$ Integrate by parts: Take $u=x^2$, $dv=\cos 2x\,dx$. Then $$du=2x\,dx,\qquad v=\frac12\sin 2x.$$ So $$J=\left[\frac{x^2}{2}\sin 2x\right]_0^{\pi/2}-\int_0^{\pi/2}x\sin 2x\,dx.$$ Boundary term is $0$, since $\sin 0=\sin \pi=0$. Hence $$J=-\int_0^{\pi/2}x\sin 2x\,dx.$$ Again integrate by parts: Take $u=x$, $dv=\sin 2x\,dx$. Then $$du=dx,\qquad v=-\frac12\cos 2x.$$ Thus $$\int_0^{\pi/2}x\sin 2x\,dx=\left[-\frac{x}{2}\cos 2x\right]_0^{\pi/2}+\frac12\int_0^{\pi/2}\cos 2x\,dx.$$ Now $$\left[-\frac{x}{2}\cos 2x\right]_0^{\pi/2}=-\frac{\pi/2}{2}\cos \pi-0=\frac{\pi}{4},$$ and $$\frac12\int_0^{\pi/2}\cos 2x\,dx=\frac12\left[\frac12\sin 2x\right]_0^{\pi/2}=0.$$ So $$\int_0^{\pi/2}x\sin 2x\,dx=\frac{\pi}{4},$$ therefore $$J=-\frac{\pi}{4}.$$ Hence $$48J=48\left(-\frac{\pi}{4}\right)=-12\pi.$$ 6. Therefore $$I=2\pi^3-12\pi=\pi(2\pi^2-12).$$ Comparing with $$I=\pi(\alpha\pi^2+\beta),$$ we get $$\alpha=2,\qquad \beta=-12.$$ So $$(\alpha+\beta)^2=(2-12)^2=(-10)^2=100.$$ 7. Hence the correct option is $$\boxed{\text{B: }100}.$$

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