Substitution Rule for Definite Integrals — Question 10

Question 10

Evaluate the definite integral $$\int_{0}^{1} \sqrt{1-x^2}\,\arcsin x \, dx.$$

Question 10 - Solution

This integral is interesting because it mixes an inverse trigonometric function with an algebraic expression, suggesting a trigonometric substitution.

Let $$x=\sin\theta.$$ Then $$dx=\cos\theta\,d\theta, \qquad \sqrt{1-x^2}=\cos\theta.$$

Change the limits of integration. When $x=0$, $\theta=0$. When $x=1$, $\theta=\frac{\pi}{2}$.

Substitute into the integral: $$\int_{0}^{1} \sqrt{1-x^2}\,\arcsin x \, dx = \int_{0}^{\pi/2} \theta\,\cos^2\theta\,d\theta.$$

Use the identity $$\cos^2\theta=\frac{1+\cos(2\theta)}{2}.$$

Then $$\int_{0}^{\pi/2} \theta\,\cos^2\theta\,d\theta = \frac12\int_{0}^{\pi/2} \theta\,d\theta + \frac12\int_{0}^{\pi/2} \theta\cos(2\theta)\,d\theta.$$

The first integral is $$\frac12\int_{0}^{\pi/2} \theta\,d\theta = \frac12\cdot\frac{(\pi/2)^2}{2} = \frac{\pi^2}{16}.$$

For the second integral, use integration by parts. Let $$u=\theta, \qquad dv=\cos(2\theta)\,d\theta.$$ Then $$du=d\theta, \qquad v=\frac12\sin(2\theta).$$

Thus, $$\frac12\int_{0}^{\pi/2} \theta\cos(2\theta)\,d\theta = \frac12\left[ \frac12\theta\sin(2\theta) - \frac12\int \sin(2\theta)\,d\theta \right]_{0}^{\pi/2}.$$

Since $\sin(\pi)=0$ and $\sin(0)=0$, the boundary term vanishes. Also, $$\int \sin(2\theta)\,d\theta=-\frac12\cos(2\theta).$$

Therefore, $$\frac12\int_{0}^{\pi/2} \theta\cos(2\theta)\,d\theta = \frac18\bigl[\cos(2\theta)\bigr]_{0}^{\pi/2} = \frac18(-1-1) = -\frac14.$$

Combine results: $$\frac{\pi^2}{16}-\frac14.$$

$$\boxed{\frac{\pi^2}{16}-\frac14}$$