Evaluate: _0^ 4 ^4 x d x - Vidyadip

Question

Evaluate: $\int_0^{\frac{\pi}{4}} \sec ^4 x d x$

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Answer

$\text { Let } I =\int_0^{\frac{\pi}{4}} \sec ^4 x d x$
$=\int_0^{\frac{\pi}{4}} \sec ^2 x \cdot \sec ^2 x d x$
$=\int_0^{\frac{\pi}{4}}\left(1+\tan ^2 x\right) \sec ^2 x d x$
Put $\tan x=t$
$\therefore \sec ^2 x d dx = dt$
When $x =0, t =0$ and when $x =\frac{\pi}{4}, t =1$
$\therefore I =\int_0^1\left(1+ t ^2\right) dt$
$=\int_0^1 dt +\int_0^1 t ^2 dt$
$=[ t ]_0^1+\left[\frac{ t ^3}{3}\right]_0^1$
$=(1-0)+\frac{1}{3}\left(1^3-0\right)$
$=\frac{4}{3}$

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