If \int\limits_-\pi / 2^\pi / 2 \frac8 \sqrt2 \cos x ~d x(1+e^\sin x)(1+\sin ^4 x)=\alpha \pi+\beta \log _e(3+2 \sqrt2) , where \alpha, \beta are integers, then \alpha^2+\beta^2 equals :

Definite Integration PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q58 with Solution

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Correct answer: 8

Let $I=\int_{-\pi/2}^{\pi/2} \frac{8\sqrt2\cos x\,dx}{(1+e^{\sin x})(1+\sin^4 x)}.$ We need to write it in the form $I=\alpha\pi+\beta\ln(3+2\sqrt2),$ where $\alpha,\beta\in\mathbb Z$ .
Use the standard property for integrals over symmetric limits: if $I=\int_{-a}^{a} f(x)\,dx,$ then $I=\int_{-a}^{a} \frac{f(x)+f(-x)}{2}\,dx.$

Here, $f(x)=\frac{8\sqrt2\cos x}{(1+e^{\sin x})(1+\sin^4 x)}.$ Now, $f(-x)=\frac{8\sqrt2\cos x}{(1+e^{-\sin x})(1+\sin^4 x)}.$ So, \begin{align*} f(x)+f(-x) &=\frac{8\sqrt2\cos x}{1+\sin^4 x}\left(\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right). \end{align*} But $\frac{1}{1+e^t}+\frac{1}{1+e^{-t}}=1.$ Hence, $f(x)+f(-x)=\frac{8\sqrt2\cos x}{1+\sin^4 x}.$ Therefore,

=4\sqrt2\int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+\sin^4 x}\,dx.$$ 3. Substitute $$t=\sin x,\qquad dt=\cos x\,dx.$$ As $x$ goes from $-\pi/2$ to $\pi/2$, $t$ goes from $-1$ to $1$. Thus, $$I=4\sqrt2\int_{-1}^{1} \frac{dt}{1+t^4}.$$ Since $\frac{1}{1+t^4}$ is even, $$I=8\sqrt2\int_0^1 \frac{dt}{1+t^4}.$$ 4. Factorize: $$t^4+1=(t^2+\sqrt2 t+1)(t^2-\sqrt2 t+1).$$ Use partial fractions: $$\frac{1}{t^4+1}=\frac{At+B}{t^2+\sqrt2 t+1}+\frac{Ct+D}{t^2-\sqrt2 t+1}.$$ On solving, we get the standard decomposition $$\frac{1}{t^4+1}=\frac{1}{2\sqrt2}\left(\frac{t+\sqrt2}{t^2+\sqrt2 t+1}-\frac{t-\sqrt2}{t^2-\sqrt2 t+1}\right).$$ Hence, \begin{align*} \int_0^1 \frac{dt}{1+t^4} &=\frac{1}{2\sqrt2}\int_0^1 \left(\frac{t+\sqrt2}{t^2+\sqrt2 t+1}-\frac{t-\sqrt2}{t^2-\sqrt2 t+1}\right)dt. \end{align*} This integrates to the known result $$\int \frac{dt}{1+t^4} =\frac{1}{4\sqrt2}\ln\left(\frac{t^2+\sqrt2 t+1}{t^2-\sqrt2 t+1}\right)+\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{\sqrt2 t}{1-t^2}\right)+C.$$ 5. Evaluate from $0$ to $1$: \begin{align*} \int_0^1 \frac{dt}{1+t^4} &=\left[\frac{1}{4\sqrt2}\ln\left(\frac{t^2+\sqrt2 t+1}{t^2-\sqrt2 t+1}\right) +\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{\sqrt2 t}{1-t^2}\right)\right]_0^1. \end{align*} At $t=1$, $$\frac{t^2+\sqrt2 t+1}{t^2-\sqrt2 t+1}=\frac{2+\sqrt2}{2-\sqrt2}=3+2\sqrt2,$$ and $$\tan^{-1}\left(\frac{\sqrt2}{0^+}\right)=\frac\pi2.$$ At $t=0$, both terms are $0$. So, $$\int_0^1 \frac{dt}{1+t^4}=\frac{1}{4\sqrt2}\ln(3+2\sqrt2)+\frac{\pi}{4\sqrt2}.$$ 6. Therefore, \begin{align*} I&=8\sqrt2\left(\frac{1}{4\sqrt2}\ln(3+2\sqrt2)+\frac{\pi}{4\sqrt2}\right)\\ &=2\ln(3+2\sqrt2)+2\pi. \end{align*} Thus, $$\alpha=2,\qquad \beta=2.$$ So, $$\alpha^2+\beta^2=2^2+2^2=8.$$ 7. Comparison with stored answer: Derived answer $=8$, which matches the stored correct answer.

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View written solutionFree

Correct answer: 8

Let $I=\int_{-\pi/2}^{\pi/2} \frac{8\sqrt2\cos x\,dx}{(1+e^{\sin x})(1+\sin^4 x)}.$ We need to write it in the form $I=\alpha\pi+\beta\ln(3+2\sqrt2),$ where $\alpha,\beta\in\mathbb Z$ .
Use the standard property for integrals over symmetric limits: if $I=\int_{-a}^{a} f(x)\,dx,$ then $I=\int_{-a}^{a} \frac{f(x)+f(-x)}{2}\,dx.$

=4\sqrt2\int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+\sin^4 x}\,dx.$$ 3. Substitute $$t=\sin x,\qquad dt=\cos x\,dx.$$ As $x$ goes from $-\pi/2$ to $\pi/2$, $t$ goes from $-1$ to $1$. Thus, $$I=4\sqrt2\int_{-1}^{1} \frac{dt}{1+t^4}.$$ Since $\frac{1}{1+t^4}$ is even, $$I=8\sqrt2\int_0^1 \frac{dt}{1+t^4}.$$ 4. Factorize: $$t^4+1=(t^2+\sqrt2 t+1)(t^2-\sqrt2 t+1).$$ Use partial fractions: $$\frac{1}{t^4+1}=\frac{At+B}{t^2+\sqrt2 t+1}+\frac{Ct+D}{t^2-\sqrt2 t+1}.$$ On solving, we get the standard decomposition $$\frac{1}{t^4+1}=\frac{1}{2\sqrt2}\left(\frac{t+\sqrt2}{t^2+\sqrt2 t+1}-\frac{t-\sqrt2}{t^2-\sqrt2 t+1}\right).$$ Hence, \begin{align*} \int_0^1 \frac{dt}{1+t^4} &=\frac{1}{2\sqrt2}\int_0^1 \left(\frac{t+\sqrt2}{t^2+\sqrt2 t+1}-\frac{t-\sqrt2}{t^2-\sqrt2 t+1}\right)dt. \end{align*} This integrates to the known result $$\int \frac{dt}{1+t^4} =\frac{1}{4\sqrt2}\ln\left(\frac{t^2+\sqrt2 t+1}{t^2-\sqrt2 t+1}\right)+\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{\sqrt2 t}{1-t^2}\right)+C.$$ 5. Evaluate from $0$ to $1$: \begin{align*} \int_0^1 \frac{dt}{1+t^4} &=\left[\frac{1}{4\sqrt2}\ln\left(\frac{t^2+\sqrt2 t+1}{t^2-\sqrt2 t+1}\right) +\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{\sqrt2 t}{1-t^2}\right)\right]_0^1. \end{align*} At $t=1$, $$\frac{t^2+\sqrt2 t+1}{t^2-\sqrt2 t+1}=\frac{2+\sqrt2}{2-\sqrt2}=3+2\sqrt2,$$ and $$\tan^{-1}\left(\frac{\sqrt2}{0^+}\right)=\frac\pi2.$$ At $t=0$, both terms are $0$. So, $$\int_0^1 \frac{dt}{1+t^4}=\frac{1}{4\sqrt2}\ln(3+2\sqrt2)+\frac{\pi}{4\sqrt2}.$$ 6. Therefore, \begin{align*} I&=8\sqrt2\left(\frac{1}{4\sqrt2}\ln(3+2\sqrt2)+\frac{\pi}{4\sqrt2}\right)\\ &=2\ln(3+2\sqrt2)+2\pi. \end{align*} Thus, $$\alpha=2,\qquad \beta=2.$$ So, $$\alpha^2+\beta^2=2^2+2^2=8.$$ 7. Comparison with stored answer: Derived answer $=8$, which matches the stored correct answer.

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