Find the integral of the function sin ^{-1}(c | vedclass

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

Question

(A) We know that $\sin ^{-1}(\cos x) = \sin ^{-1}(\sin(\frac{\pi}{2}-x))$.Since $\sin ^{-1}(\sin 	heta) = 	heta$ for $	heta \in [-\frac{\pi}{2}, \f

Vedclass · Accepted Answer

$\frac{\pi x}{2}-\frac{x^{2}}{2}+C_{1}$

Vedclass · Answer

(A) We know that $\sin ^{-1}(\cos x) = \sin ^{-1}(\sin(\frac{\pi}{2}-x))$.Since $\sin ^{-1}(\sin 	heta) = 	heta$ for $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin ^{-1}(\cos x) = \frac{\pi}{2}-x$.Now,we integrate the function:$\int \sin ^{-1}(\cos x) \, dx = \int (\frac{\pi}{2}-x) \, dx$$= \int \frac{\pi}{2} \, dx - \int x \, dx$$= \frac{\pi}{2}x - \frac{x^{2}}{2} + C_{1}$.

Find the integral of the function $\sin ^{-1}(\cos x)$.

Similar Questions

Find the integral of the function $\sin 4x \sin 8x$.

If $f'(x) = x^2 + 5$ and $f(0) = -1$,then $f(x) = $

$\int \sec \left(x-\frac{\pi}{3}\right) \sec \left(x+\frac{\pi}{6}\right) d x=$

$\int \frac{1}{1+x+x^2} \, dx =$

If $\int \frac{x^4+1}{x^2+1} dx = Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E$,then $A+B+C+D=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module