If the value of the integral \int_-\frac\pi2^\frac\pi2(\fracx^2 \cos x1+\pi^x+\frac1+\sin ^2 x1+e^\sin x^2123) d x=\frac\pi4(\pi+a)-2 , then the value of a is

Definite Integration PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q50 with Solution

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Correct answer: B

Let $I=\int_{-\pi/2}^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{1+\sin^2 x}{1+e^{\sin(x^{2123})}}\right)dx.$ We evaluate the two parts separately.

First integral: $I_1=\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+\pi^x}\,dx.$

Use the standard property $\int_{-a}^{a} f(x)\,dx=\int_{0}^{a}[f(x)+f(-x)]dx.$ So, $I_1=\int_0^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{x^2\cos x}{1+\pi^{-x}}\right)dx.$ Now, $\frac{1}{1+\pi^x}+\frac{1}{1+\pi^{-x}}=1,$ because $\frac{1}{1+\pi^{-x}}=\frac{\pi^x}{1+\pi^x}.$ Hence, $I_1=\int_0^{\pi/2}x^2\cos x\,dx.$

Now integrate by parts: $\int x^2\cos x\,dx=x^2\sin x-\int 2x\sin x\,dx.$ Again, $\int x\sin x\,dx=-x\cos x+\sin x.$ Therefore, $\int x^2\cos x\,dx=x^2\sin x+2x\cos x-2\sin x.$ Evaluating from $0$ to $\pi/2$ :

=\frac{\pi^2}{4}-2.$$ --- 3. Second integral: $$I_2=\int_{-\pi/2}^{\pi/2}\frac{1+\sin^2 x}{1+e^{\sin(x^{2123})}}dx.$$ Since $2123$ is odd, $$(-x)^{2123}=-x^{2123}.$$ Also $\sin$ is odd, so $$\sin((-x)^{2123})=\sin(-x^{2123})=-\sin(x^{2123}).$$ Let $$g(x)=\frac{1+\sin^2 x}{1+e^{\sin(x^{2123})}}.$$ Then $$g(-x)=\frac{1+\sin^2 x}{1+e^{-\sin(x^{2123})}}.$$ So, $$g(x)+g(-x)=(1+\sin^2 x)\left(\frac{1}{1+e^u}+\frac{1}{1+e^{-u}}\right),$$ where $u=\sin(x^{2123})$. But $$\frac{1}{1+e^u}+\frac{1}{1+e^{-u}}=1.$$ Hence, $$g(x)+g(-x)=1+\sin^2 x.$$ Therefore, $$I_2=\int_0^{\pi/2}(1+\sin^2 x)dx.$$ Now, $$\int_0^{\pi/2}1\,dx=\frac{\pi}{2},$$ and $$\int_0^{\pi/2}\sin^2 x\,dx=\frac{\pi}{4}.$$ Thus, $$I_2=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}.$$ --- 4. Total integral: $$I=I_1+I_2=\left(\frac{\pi^2}{4}-2\right)+\frac{3\pi}{4} =\frac{\pi}{4}(\pi+3)-2.$$ Given, $$I=\frac{\pi}{4}(\pi+a)-2.$$ Comparing, $$a=3.$$ --- 5. Checking options: - A: $-\frac{3}{2}$ ❌ - B: $3$ ✅ - C: $\frac{3}{2}$ ❌ - D: $2$ ❌ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Let $I=\int_{-\pi/2}^{\pi/2}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{1+\sin^2 x}{1+e^{\sin(x^{2123})}}\right)dx.$ We evaluate the two parts separately.

First integral: $I_1=\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{1+\pi^x}\,dx.$

=\frac{\pi^2}{4}-2.$$ --- 3. Second integral: $$I_2=\int_{-\pi/2}^{\pi/2}\frac{1+\sin^2 x}{1+e^{\sin(x^{2123})}}dx.$$ Since $2123$ is odd, $$(-x)^{2123}=-x^{2123}.$$ Also $\sin$ is odd, so $$\sin((-x)^{2123})=\sin(-x^{2123})=-\sin(x^{2123}).$$ Let $$g(x)=\frac{1+\sin^2 x}{1+e^{\sin(x^{2123})}}.$$ Then $$g(-x)=\frac{1+\sin^2 x}{1+e^{-\sin(x^{2123})}}.$$ So, $$g(x)+g(-x)=(1+\sin^2 x)\left(\frac{1}{1+e^u}+\frac{1}{1+e^{-u}}\right),$$ where $u=\sin(x^{2123})$. But $$\frac{1}{1+e^u}+\frac{1}{1+e^{-u}}=1.$$ Hence, $$g(x)+g(-x)=1+\sin^2 x.$$ Therefore, $$I_2=\int_0^{\pi/2}(1+\sin^2 x)dx.$$ Now, $$\int_0^{\pi/2}1\,dx=\frac{\pi}{2},$$ and $$\int_0^{\pi/2}\sin^2 x\,dx=\frac{\pi}{4}.$$ Thus, $$I_2=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}.$$ --- 4. Total integral: $$I=I_1+I_2=\left(\frac{\pi^2}{4}-2\right)+\frac{3\pi}{4} =\frac{\pi}{4}(\pi+3)-2.$$ Given, $$I=\frac{\pi}{4}(\pi+a)-2.$$ Comparing, $$a=3.$$ --- 5. Checking options: - A: $-\frac{3}{2}$ ❌ - B: $3$ ✅ - C: $\frac{3}{2}$ ❌ - D: $2$ ❌ So the correct option is **B**.

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