Question

Evaluate : \( \displaystyle\int_{0}^{1} x \sin^{-1}x \, dx \)

Hint:

Use integration by parts with u = sin^{-1}x and dv = x dx, or put x = sin(theta). The answer is pi/8.

Answer 1

Step 1: Choose the method:
The integrand is a product of the algebraic function \( x \) and the inverse trigonometric function \( \sin^{-1}x \).
So we use integration by parts, taking \( u = \sin^{-1}x \) (the inverse function goes first) and \( dv = x\,dx \).
Then \( du = \dfrac{dx}{\sqrt{1-x^2}} \) and \( v = \dfrac{x^2}{2} \).

Step 2: Apply the formula:
We use \( \int u\,dv = uv - \int v\,du \).
\[ \int_0^1 x\sin^{-1}x\,dx = \left[ \frac{x^2}{2}\sin^{-1}x \right]_0^1 - \frac{1}{2}\int_0^1 \frac{x^2}{\sqrt{1-x^2}}\,dx \]

Step 3: Evaluate the boundary term:
At \( x = 1 \) the term is \( \dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi}{4} \).
At \( x = 0 \) the term is \( 0 \).
So the boundary term equals \( \dfrac{\pi}{4} \).

Step 4: Evaluate the remaining integral:
Let \( J = \displaystyle\int_0^1 \frac{x^2}{\sqrt{1-x^2}}\,dx \). Put \( x = \sin\theta \), so \( dx = \cos\theta\,d\theta \).
When \( x = 0 \), \( \theta = 0 \). When \( x = 1 \), \( \theta = \dfrac{\pi}{2} \). Also \( \sqrt{1-x^2} = \cos\theta \).
\[ J = \int_0^{\pi/2} \sin^2\theta\,d\theta = \int_0^{\pi/2} \frac{1-\cos 2\theta}{2}\,d\theta = \left[ \frac{\theta}{2} - \frac{\sin 2\theta}{4} \right]_0^{\pi/2} = \frac{\pi}{4} \]

Step 5: Combine the two parts:
\[ \int_0^1 x\sin^{-1}x\,dx = \frac{\pi}{4} - \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8} \]

Final Answer:
The value of the integral is \( \dfrac{\pi}{8} \). \[ \boxed{\dfrac{\pi}{8}} \]