Question

Find \(\displaystyle\int\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx\).

Hint:

Use integration by parts with \(u=\sin^{-1}x\) and \(dv=\dfrac{x}{\sqrt{1-x^2}}dx\), so \(v=-\sqrt{1-x^2}\).

Answer 1

Step 1: Understanding the Concept:
The integral \(\displaystyle\int\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}dx\) has a product of \(\sin^{-1}x\) and a function of x, so integration by parts is used.
Choose \(u=\sin^{-1}x\) (easy to differentiate) and \(dv=\dfrac{x}{\sqrt{1-x^2}}dx\) (easy to integrate).

Step 2: Find v and du:
For dv, let \(w=1-x^2\), \(dw=-2x\,dx\).
\[ v=\int\frac{x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2} \]
\[ du=\frac{1}{\sqrt{1-x^2}}dx \]

Step 3: Apply integration by parts:
Use \(\displaystyle\int u\,dv=uv-\int v\,du\).
\[ \int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}\sin^{-1}x-\int\left(-\sqrt{1-x^2}\right)\frac{1}{\sqrt{1-x^2}}dx \]
\[ =-\sqrt{1-x^2}\sin^{-1}x+\int1\,dx \]

Step 4: Finish the integration:
\[ \int1\,dx=x \]

Final Answer:
Combine both parts to get the antiderivative. \[ \boxed{\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx=x-\sqrt{1-x^2}\sin^{-1}x+C} \]