assignment submission for this assignment, you submit answers by question parts. the number of subm…

Answer:

$x\sin^{- 1}(x)+\sqrt{1 - x^{2}}+C$

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \sin^{-1}(x)$ and $dv=dx$. Then $du=\frac{1}{\sqrt{1 - x^{2}}}dx$ and $v = x$.

Step2: Apply the formula

$\int\sin^{-1}(x)dx=x\sin^{-1}(x)-\int\frac{x}{\sqrt{1 - x^{2}}}dx$.

Step3: Solve the new integral

Let $t = 1 - x^{2}$, then $dt=-2x;dx$ and $x;dx=-\frac{1}{2}dt$. So $\int\frac{x}{\sqrt{1 - x^{2}}}dx=-\frac{1}{2}\int t^{-\frac{1}{2}}dt$.

Step4: Integrate $-\frac{1}{2}\int t^{-\frac{1}{2}}dt$

$-\frac{1}{2}\int t^{-\frac{1}{2}}dt=-\frac{1}{2}\times\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C=-\sqrt{t}+C=-\sqrt{1 - x^{2}}+C$.

Step5: Combine the results

$\int\sin^{-1}(x)dx=x\sin^{-1}(x)-(-\sqrt{1 - x^{2}})+C=x\sin^{-1}(x)+\sqrt{1 - x^{2}}+C$.