Verified Solution • 2025–26 Session moderate

QJEEM24Ma3737: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3737 Solution: If , where are integers, then equals :

If ∫−π/2π/282cos⁡x dx(1+esin⁡x)(1+sin⁡4x)=απ+βlog⁡e(3+22)\int\limits_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \mathrm{~d} x}{\left(1+\mathrm{e}^{\sin x}\right)\left(1+\sin ^4 x\right)}=\alpha \pi+\beta \log _{\mathrm{e}}(3+2 \sqrt{2}) , where α,β\alpha, \beta are integers, then α2+β2\alpha^2+\beta^2 equals :

Step-by-Step Verified Solution

I=∫−π2π282cos⁡x(1+esin⁡x)(1+sin⁡4x)dxI=\int\limits_{-\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 ............(1)

Apply king's rule

I=∫−π2π282cos⁡x(esin⁡x)(1+esin⁡x)(1+sin⁡4x)dxI=\int\limits_{-\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 ..........(2)

Adding (1) and (2), we get

amp;2I=∫−π2π282cos⁡x1+sin⁡4xdxamp;I=∫0π282cos⁡x1+sin⁡4xdx,\begin{aligned}& 2 I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x \\\\& I=\int\limits_0^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x,\end{aligned}

Let sin⁡x=t\sin x=t

amp;I=82∫01dt1+t4amp;=42∫01(1+1t2)−(1−1t2)t2+1t2dt\begin{aligned}& I=8 \sqrt{2} \int\limits_0^1 \frac{d t}{1+t^4} \\\\& =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t\end{aligned}

amp;=42∫01(1+1t2)dt(t−1t2)2+2−42∫01(1−1t2)dt(t+1t2)2−2amp;=42⋅12(tan⁡−1t−1t2∣01−42⋅122[log⁡∣t+1t−2t+1t+2∣]01amp;=2π−2log⁡∣2−22+2∣\begin{aligned} & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t^2}\right)^2+2}-4 \sqrt{2} \int\limits_0^1 \frac{\left(1-\frac{1}{t^2}\right) d t}{\left(t+\frac{1}{t^2}\right)^2-2} \\\\ & =4 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\left(\left.\tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}\right|_0 ^1-4 \sqrt{2} \cdot \frac{1}{2 \sqrt{2}}\left[\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|\right]_0^1\right. \\\\ & =2 \pi-2 \log \left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right|\end{aligned}

amp;=2π+2log⁡(3+22)=απ+βlog⁡e(3+22)amp;⇒α=2,β=2amp;⇒α2+β2=8\begin{aligned} & =2 \pi+2 \log (3+2 \sqrt{2})=\alpha \pi+\beta \log _e(3+2 \sqrt{2}) \\\\ & \Rightarrow \alpha=2, \beta=2 \\\\ & \Rightarrow \alpha^2+\beta^2=8\end{aligned}

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