( + ) 2 d = - Vidyadip

MCQ

$\int {\frac{{(\sin \theta + \cos \theta )}}{{\sqrt {\sin 2\theta } }}} d\theta = $

A
$\log \left| {\cos \theta - \sin \theta + \sqrt {\sin 2\theta } } \right|$
B
$\log \left| {\sin \theta - \cos \theta ) + \sqrt {\sin 2\theta } } \right|$
✓
${\sin ^{ - 1}}(\sin \theta - \cos \theta ) + c$
D
${\sin ^{ - 1}}(\sin \theta + \cos \theta ) + c$

✓

Answer

Correct option: C.

${\sin ^{ - 1}}(\sin \theta - \cos \theta ) + c$

(c) Let $I = \int {\frac{{\sin \theta + \cos \theta }}{{\sqrt {2\sin \theta \cos \theta } }}d\theta } $
$I = \int {\frac{{\sin \theta + \cos \theta }}{{\sqrt {1 - (1 - 2\sin \theta \cos \theta )} }}d\theta } $
$ = \int {\frac{{(\sin \theta + \cos \theta )d\theta }}{{\sqrt {1 - ({{\sin }^2}\theta + {{\cos }^2}\theta - 2\sin \theta \cos \theta )} }}} $
$ = \int {\frac{{\sin \theta + \cos \theta }}{{\sqrt {1 - {{(\sin \theta - \cos \theta )}^2}} }}d\theta } $
Let $(\sin \theta - \cos \theta ) = t$ ==> $(\cos \theta + \sin \theta )d\theta = dt$
$I = \int {\frac{{dt}}{{\sqrt {1 - {t^2}} }} = {{\sin }^{ - 1}}(t) + c} = {\sin ^{ - 1}}(\sin \theta - \cos \theta ) + c$.

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