1. Tardigrade
  2. Question
  3. Mathematics
  4. The value of the integral ∫ x sin -1 x d x is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The value of the integral $\int x \sin ^{-1} x d x$ is equal to

Integrals

Solution:

Let $I=\int x \sin ^{-1} x d x$
$\Rightarrow I=\left(\sin ^{-1} x \frac{x^2}{2}-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} d x\right.$
$\Rightarrow I=\frac{x^2}{2} \sin ^{-1} x+\frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} d x$
$=\frac{x^2}{2} \sin ^{-1} x+\frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} d x$
$\Rightarrow =\frac{x^2}{2} \sin ^{-1} x+\frac{1}{2}\left\{\int \frac{1-x^2}{\sqrt{1-x^2}} d x-\int \frac{1}{\sqrt{1-x^2}} d x\right\}$
$\Rightarrow I=\frac{x^2}{2} \sin ^{-1} x+\frac{1}{2}\left\{\int \sqrt{1-x^2} d x-\int \frac{1}{\sqrt{1-x^2}} d x\right\}$
$=\frac{x^2}{2} \sin ^{-1} x+\frac{1}{2}\left[\left\{\frac{1}{2} x \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right\}-\sin ^{-1} x\right] +C$
$\Rightarrow I=\frac{1}{2} x^2 \sin ^{-1} x+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{4} \sin ^{-1} x+C$