MCQ
$\int {\sqrt {({{\sin }^2}x)} } \,\,dx = \,;\,(x \ne n\pi ,n \in I)$
  • A
    $-\cos\, x + c$
  • B
    $\cos \,x + c$
  • $-\cos\, x \,sgn. (\sin \,x) + c$
  • D
    None of these

Answer

Correct option: C.
$-\cos\, x \,sgn. (\sin \,x) + c$
c
$\int|\sin x| d x$

$=\left\{\begin{array}{l}{\int \sin x d x ; \sin x \geq 0} \\ {-\int \sin x d x ; \sin x<0}\end{array}\right.$

$ = \left\{ {\begin{array}{*{20}{c}}
{\int  -  \cos x;}&{\sin x \ge 0}\\
{\cos x;}&{\sin x < 0}
\end{array}} \right.$

$ =  - \cos x{\mathop{\rm \,\,sgn}} \,(\sin x) + c$

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