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MCQ

$\int_{}^{} {\frac{{d\theta }}{{\sin \theta {{\cos }^3}\theta }} = } $

A
$\log \tan \theta + {\tan ^2}\theta + c$
B
$\log \tan \theta - \frac{1}{2}{\tan ^2}\theta + c$
✓
$\log \tan \theta + \frac{1}{2}{\tan ^2}\theta + c$
D
None of these

✓

Answer

Correct option: C.

$\log \tan \theta + \frac{1}{2}{\tan ^2}\theta + c$

c
(c)$\int_{}^{} {\frac{{d\theta }}{{\sin \theta {{\cos }^3}\theta }} = \int_{}^{} {\frac{{{{\sec }^2}\theta \,d\theta }}{{\sin \theta \cos \theta }} = \int_{}^{} {\frac{{{{\sec }^2}\theta (1 + {{\tan }^2}\theta )}}{{\tan \theta }}} } } {\rm{ }}d\theta $
Put $t = \tan \theta \Rightarrow dt = {\sec ^2}\theta \,d\theta ,$ then it reduces to
$\int_{}^{} {\frac{{1 + {t^2}}}{t}\,dt = \int_{}^{} {\left( {\frac{1}{t} + t} \right)\,dt} } $
$ = \log t + \frac{{{t^2}}}{2} + c = \log \tan \theta + \frac{{{{\tan }^2}\theta }}{2} + c.$

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