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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

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

Similar Questions

The value of $\int {\frac{1}{{{{(x - 5)}^2}}}\,dx} $ is

$\int {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx} = $

The value of $\int \frac{\sin x}{\cos^2 x} \, dx$ is

If $\int \frac{dx}{\sqrt{16-9x^2}} = A \sin^{-1}(Bx) + C$,then $A+B=$

Integrate the function: $\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module