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

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

Question

(B) Let $I = \int \frac{\sin ^{-1} \sqrt{x} - \cos ^{-1} \sqrt{x}}{\sqrt{x}(\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x})} dx$. Since $\sin ^{-1} \sqrt{x

Vedclass · Accepted Answer

$\frac{8}{\pi}\left(\sqrt{x} \sin ^{-1} \sqrt{x}+\sqrt{1-x}\right)-2 \sqrt{x}+C$

Vedclass · Answer

(B) Let $I = \int \frac{\sin ^{-1} \sqrt{x} - \cos ^{-1} \sqrt{x}}{\sqrt{x}(\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x})} dx$. Since $\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x} = \frac{\pi}{2}$,we have $\cos ^{-1} \sqrt{x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x}$. Substituting this,$I = \int \frac{\sin ^{-1} \sqrt{x} - (\frac{\pi}{2} - \sin ^{-1} \sqrt{x})}{\sqrt{x}(\frac{\pi}{2})} dx = \frac{2}{\pi} \int \frac{2 \sin ^{-1} \sqrt{x} - \frac{\pi}{2}}{\sqrt{x}} dx$. Let $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2 dt$. $I = \frac{2}{\pi} \int (2 \sin ^{-1} t - \frac{\pi}{2}) (2 dt) = \frac{8}{\pi} \int \sin ^{-1} t dt - 2 \int dt$. Using $\int \sin ^{-1} t dt = t \sin ^{-1} t + \sqrt{1-t^2} + C$,we get: $I = \frac{8}{\pi} (t \sin ^{-1} t + \sqrt{1-t^2}) - 2t + C$. Substituting $t = \sqrt{x}$: $I = \frac{8}{\pi} (\sqrt{x} \sin ^{-1} \sqrt{x} + \sqrt{1-x}) - 2\sqrt{x} + C$.

$\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sqrt{x}\left(\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}\right)} d x=$

Similar Questions

$\int {e^{\log (\sin x)}} \, dx = $

$\int \frac{1}{\cos x+\sqrt{3} \sin x} dx =$

The integral of $\sqrt{1 + 2 \cot x (\cot x + \csc x)}$ with respect to $x$ is:

If $\int \frac{dx}{32-2x^2} = A \log(4-x) + B \log(4+x) + c$,then the values of $A$ and $B$ are respectively (where $c$ is a constant of integration).

$ \int \frac{1}{1+e^{x}} d x $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module