If I=2/π int -π / 4 π / 4 frac d x (1+e sin x )(2-cos 2 x) then 27 I 2 equals \ \ \ \

Mathematics · Definite Integration

JEE Advanced 2019 — Paper 1 — Question 42

If $I=\frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{d x}{\left(1+e^{\sin x}\right)(2-\cos 2 x)}$ then $27 I^{2}$ equals $\_\_\_\_$

Numerical answer — enter this value.

\begin{gathered} I=\frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{d x}{\left(1+e^{\sin x}\right)(2-\cos 2 x)} …(i)\end{gathered}

$\mathrm{x}=-\mathrm{t}$

\begin{gathered} \mathrm{I}=\frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{\mathrm{e}^{\sin \mathrm{t}}}{\left(1+\mathrm{e}^{\sin t}\right)(2-\cos 2 \mathrm{t})} \mathrm{dt} …(ii) \end{gathered}

add (i) and (ii)

\begin{aligned} & \mathrm{I}=\frac{1}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\mathrm{x} \cos 2 \mathrm{t}} \mathrm{dt}=\frac{1}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{\sec ^{2} \mathrm{t}}{1+3 \tan ^{2} \mathrm{t}} \mathrm{dt}=\frac{1}{\pi} \cdot \frac{2 \pi}{3 \sqrt{3}} \\& 3 \sqrt{3} \mathrm{I}=2 \Rightarrow 27 \mathrm{I}^{2}=4 \end{aligned}

Answer key and solution verified before publishing.

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