The value of _ ^ x ^2 x ;dx is - Vidyadip

MCQ

The value of $\int_{}^{} {\frac{{\sin x}}{{{{\cos }^2}x}}\;dx} $ is

A
$\sin x + k$
B
$\tan x + k$
✓
$\sec x + k$
D
$\tan x + \sec x + k$

✓

Answer

Correct option: C.

$\sec x + k$

c
(c) Given $I = \int_{}^{} {\frac{{\sin x}}{{{{\cos }^2}x}}} \,dx$.

Put $\cos x = t \Rightarrow \sin x\,dx = - dt$
$\therefore \,\,\,I = \int_{}^{} {\frac{{ - dt}}{{{t^2}}}} = \frac{1}{t} + k = \sec x + k$.

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