JEE Main 2024 — 1 February, Shift 1 — Question 28

$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{\left(1+e^{\sin x}\right)\left(1+\sin ^{4} x\right)} d x$

Apply king $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+\sin ^{4} x\right)} d x$ adding (1) & (2)

$2 I=\int_{-\pi}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^{4} x} d x$ $I=\int_{0}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^{4} x} d x$

$\sin x=t$ $I=\int_{0}^{1} \frac{8 \sqrt{2}}{1+t^{4}} d x$

$I=4 \sqrt{2} \int_{0}^{1}\left(\frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}-\frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}\right) d t$

$I=4 \sqrt{2} \int_{0}^{1} \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2}-\frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t+\frac{1}{t}\right)^{2}-2} d t$

Let $t-\frac{1}{t}=z \& t+\frac{1}{t}=k$

$=4 \sqrt{2}\left[\int_{-\infty}^{0} \frac{d z}{z^{2}+2}-\int_{\infty}^{2} \frac{d k}{k^{2}-2}\right]$

$=4 \sqrt{2}\left[\frac{1}{\sqrt{2}} \tan ^{-1} \frac{Z}{\sqrt{2}}\right]_{-\infty}^{0}-\left[\frac{1}{2 \sqrt{2}} \ln \left(\frac{k-\sqrt{2}}{k+\sqrt{2}}\right)\right]_{\infty}^{2}$

$=4 \sqrt{2}\left[\frac{\pi}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\left[\ln \frac{2-\sqrt{2}}{2+\sqrt{2}}\right]\right]$

$=2 \pi+2 \ln (3+2 \sqrt{2})$

$\alpha=2$

$\beta=2$

$\alpha^{2}+\beta^{2}=8$