int {sqrt {{{sin }^2}x} } ,,dx = ,;,(x ne npi | vedclass

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Question

(C) We know that $\sqrt{\sin^2 x} = |\sin x|$.So,the integral becomes $\int |\sin x| dx$.By the definition of the absolute value function,we have:$\in

Vedclass · Accepted Answer

$-\cos x \cdot 	ext{sgn}(\sin x) + c$

Vedclass · Answer

(C) We know that $\sqrt{\sin^2 x} = |\sin x|$.So,the integral becomes $\int |\sin x| dx$.By the definition of the absolute value function,we have:$\int |\sin x| dx = \begin{cases} \int \sin x dx & 	ext{if } \sin x \ge 0 \ -\int \sin x dx & 	ext{if } \sin x Evaluating these integrals:If $\sin x \ge 0$,the integral is $-\cos x + c$.If $\sin x This can be written compactly using the signum function as $-\cos x \cdot 	ext{sgn}(\sin x) + c$.

$\int {\sqrt {{{\sin }^2}x} } \,\,dx = \,;\,(x \ne n\pi ,n \in I)$

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