_0^ /6 x ^3 x ,dx = - Vidyadip

MCQ

$\int_0^{\pi /6} {\frac{{\sin x}}{{{{\cos }^3}x}}\,dx = } $

A
$\frac{2}{3}$
✓
$\frac{1}{6}$
C
$2$
D
$\frac{1}{3}$

✓

Answer

Correct option: B.

$\frac{1}{6}$

b
(b) Let $I = \int_0^{\pi /6} {\frac{{\sin x}}{{{{\cos }^3}x}}dx = \int_0^{\pi /6} {\,\tan x{{\sec }^2}x\,dx} } $

Put $t = \tan x $

$\Rightarrow dt = {\sec ^2}x\,dx,$ then we have

$I = \int_0^{\frac{1}{{\sqrt 3 }}} {t\,dt = } \left[ {\frac{{{t^2}}}{2}} \right]_0^{\frac{1}{{\sqrt 3 }}} = \frac{1}{6}$.

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