int {frac{{{{sin }^{ - 1}}x - {{cos }^{ - 1}} | vedclass

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

Question

(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Substituting this into the integral: $I =

Vedclass · Accepted Answer

$\frac{4}{\pi }\left( {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right) - x + c$

Vedclass · Answer

(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Substituting this into the integral: $I = \int \frac{\sin^{-1} x - (\frac{\pi}{2} - \sin^{-1} x)}{\frac{\pi}{2}} dx$ $I = \frac{2}{\pi} \int (2 \sin^{-1} x - \frac{\pi}{2}) dx$ $I = \frac{4}{\pi} \int \sin^{-1} x dx - \int dx$ Using integration by parts for $\int \sin^{-1} x dx$,where $u = \sin^{-1} x$ and $dv = dx$: $\int \sin^{-1} x dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx = x \sin^{-1} x + \sqrt{1-x^2}$. Substituting this back: $I = \frac{4}{\pi} (x \sin^{-1} x + \sqrt{1-x^2}) - x + c$.

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

Similar Questions

If $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ and $f(0) = \pi$,then $f(x) =$

$\int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x=$

If $f(x) = g(x)$,then the value of $\int {f'(x) \cdot g(x)} \, dx$ is

If an antiderivative of $f(x)$ is $e^x$ and an antiderivative of $g(x)$ is $\cos x$, then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx =$

$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module