Question
$\int_0^{\frac{\pi}{4}} \frac{e^{\tan x}}{\cos ^2 x} d x=?$
✓
Answer
$(a) ( e -1)$
Explanation: $I=\int_0^{\frac{\pi}{4}} e^{\tan x} \sec ^2 x d x$
Let, $\tan x = t$,
Differentiating both side with respect to $t$
$\sec ^2 x \frac{d x}{d t}=1$
$\Rightarrow \sec ^2 x d x=d t$
$\text { At } x=0, t=0$
$\text { At } x=\frac{\pi}{4}, t=1$
$I=\int_0^1 e^t d t$
$=e^{t_0^1}$
$= e ^1- e ^0$
$= e -1$
Explanation: $I=\int_0^{\frac{\pi}{4}} e^{\tan x} \sec ^2 x d x$
Let, $\tan x = t$,
Differentiating both side with respect to $t$
$\sec ^2 x \frac{d x}{d t}=1$
$\Rightarrow \sec ^2 x d x=d t$
$\text { At } x=0, t=0$
$\text { At } x=\frac{\pi}{4}, t=1$
$I=\int_0^1 e^t d t$
$=e^{t_0^1}$
$= e ^1- e ^0$
$= e -1$
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