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Q.
Let $I=\int\limits_0^{\pi / 4} \sin 4 x \cdot e^{\tan ^2 x} d x$. Then which of the following is/are CORRECT? (where [.] denotes greatest integer function)
Integrals
Solution:
Let $\int\limits_0^{\frac{\pi}{4}} 2 \cdot \sin 2 x \cdot \cos 2 x \cdot e^{\tan ^2 x} d x=I$, then
$I=\int\limits_0^{\frac{\pi}{4}} 2 \cdot \frac{2 \tan x}{1+\tan ^2 x} \cdot \frac{1-\tan ^2 x}{1+\tan ^2 x} \cdot e^{\tan ^2 x} d x $
$I=\int\limits_0^{\frac{\pi}{4}} \frac{2\left(1-\tan ^2 x\right)}{\left(1+\tan ^2 x\right)} \cdot \frac{2 \tan x \cdot \sec ^2 x}{\left(1+\tan ^2 x\right)} \cdot e^{\tan ^2 x} d x$
$\tan ^2 x = t $
$2 \tan x \cdot \sec ^2 x d x=d t $
$\text { At } x =0, t =0$
$x=\frac{\pi}{4}, t=1 $
$I=\int\limits_0^1 \frac{2(1-t)}{(1+t)^3} \cdot e^t d t=-2 \int\limits_0^1 \frac{t-1}{(1+t)^3} \cdot e^t d t=-2 \int\limits_0^1 \frac{(t+1)-2}{(1+t)^3} \cdot e^t d t=-2 \int\limits_0^1 e^t\left[\frac{1}{(t+1)^2}+\frac{-2}{(1+t)^3}\right] d t$
$=-2\left[\frac{ e ^{ t }}{(1+ t )^2}\right]_0^1=-2\left(\frac{ e }{4}-1\right)=2-\frac{ e }{2}$