Evaluate: _0^ / 4 ^3 x d x - Vidyadip

MCQ

Evaluate: $\int_0^{\pi / 4} \tan ^3 x d x$

A
$(1-\log 2)$
B
$(1+\log 2)$
✓
$\frac{1}{2}(1-\log 2)$
D
$\frac{1}{2}(1+\log 2)$

✓

Answer

Correct option: C.

$\frac{1}{2}(1-\log 2)$

$\text { : Let } I= \int_0^{\pi / 4} \tan ^3 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) \tan x d x$
$ =\int_0^{\pi / 4} \sec ^2 x \tan x d x-\int_0^{\pi / 4} \tan x d x$
Put $\tan x=t$ in first integral
$\Rightarrow \sec ^2 x d x=d t$
When
$x=0$
$\Rightarrow t=0$
$\therefore x=\pi / 4$
$\Rightarrow t=1$
$I=\int_0^1 t d t-\int_0^{\pi / 4} \tan x d x=\left[\frac{t^2}{2}\right]_0^1-[\log |\sec x|]_0^{\pi / 4}$
$=\left(\frac{1}{2}-0\right)-\log \left|\sec \frac{\pi}{4}\right|+\log |\sec 0|=\frac{1}{2}(1-\log 2)$

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