Question

Evaluate the integral \[ \int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx . \]

Hint:

Whenever an integral contains \(\sin^{-1}x\) along with \(\frac{1}{\sqrt{1-x^2}}\), try the substitution \[ t=\sin^{-1}x \] since its derivative naturally appears in the integrand.

Answer 1

Concept: To evaluate this integral, we use the substitution method. Notice that the derivative of \( \sin^{-1}x \) is \[ \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} \] which appears in the denominator of the integrand.
Step 1: Use substitution. Let \[ t=\sin^{-1}x \] Then \[ \frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}} \] \[ dt=\frac{dx}{\sqrt{1-x^2}} \] Thus the integral becomes \[ \int x(\sin^{-1}x)\frac{dx}{\sqrt{1-x^2}} = \int x\,t\,dt \]
Step 2: Express \(x\) in terms of \(t\). Since \(t=\sin^{-1}x\), \[ x=\sin t \] Hence the integral becomes \[ \int t\sin t\,dt \]
Step 3: Apply integration by parts. Let \[ u=t, \qquad dv=\sin t\,dt \] Then \[ du=dt, \qquad v=-\cos t \] Using the formula \[ \int u\,dv = uv-\int v\,du \] \[ \int t\sin t\,dt = -t\cos t+\int \cos t\,dt \] \[ = -t\cos t+\sin t+C \]
Step 4: Substitute back \(t=\sin^{-1}x\). \[ = -(\sin^{-1}x)\cos(\sin^{-1}x)+\sin(\sin^{-1}x)+C \] Now, \[ \sin(\sin^{-1}x)=x \] and \[ \cos(\sin^{-1}x)=\sqrt{1-x^2} \] Thus, \[ \int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx = x-(\sin^{-1}x)\sqrt{1-x^2}+C \]
Step 5: Final Answer. \[ \boxed{ \int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx = x-(\sin^{-1}x)\sqrt{1-x^2}+C } \]