Question

The value of $\int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t}dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t}dt$

Answer 1

$\frac{\pi}{4}$

Answer 2

We need to evaluate

$$I = \int_{0}^{\sin^2 x} \sin^{-1}\sqrt{t}\,dt + \int_{0}^{\cos^2 x} \cos^{-1}\sqrt{t}\,dt.$$

Step 1. Differentiate $I$ with respect to $x$ using the Leibniz rule.

For the first integral:

$$\frac{d}{dx}\int_{0}^{\sin^2 x} \sin^{-1}\sqrt{t}\,dt = \sin^{-1}\sqrt{\sin^2 x}\cdot \frac{d}{dx}(\sin^2 x).$$

Since for $x\in[0,\pi/2]$, $\sqrt{\sin^2 x}=\sin x$ and $\sin^{-1}(\sin x)=x$, and

$$\frac{d}{dx}(\sin^2 x)=2\sin x\cos x = \sin 2x,$$

this term becomes:

$$x\sin2x.$$

For the second integral:

$$\frac{d}{dx}\int_{0}^{\cos^2 x} \cos^{-1}\sqrt{t}\,dt = \cos^{-1}\sqrt{\cos^2 x}\cdot \frac{d}{dx}(\cos^2 x).$$

Similarly, for $x\in[0,\pi/2]$ we have $\sqrt{\cos^2 x}=\cos x$ and $\cos^{-1}(\cos x)=x$. Also,

$$\frac{d}{dx}(\cos^2 x)= -2\sin x\cos x = -\sin2x.$$

Thus, the second term is:

$$x(-\sin2x) = -x\sin2x.$$

So the derivative of $I$ is

$$\frac{dI}{dx} = x\sin2x - x\sin2x = 0.$$

This shows that $I$ is constant (independent of $x$).

Step 2. Determine the constant by choosing an easy value of $x$. Let $x=\pi/2$.

Then, $\sin^2(\pi/2)=1$ and $\cos^2(\pi/2)=0$. Therefore,

$$I = \int_0^1 \sin^{-1}\sqrt{t}\,dt + \int_0^0 \cos^{-1}\sqrt{t}\,dt = \int_0^1 \sin^{-1}\sqrt{t}\,dt.$$

Step 3. Compute $J = \int_{0}^{1} \sin^{-1}\sqrt{t}\,dt$.

Let $t=u^2$, so that $dt=2u\,du$, and when $t=0$, $u=0$; $t=1$, $u=1$. Then,

$$J = \int_{0}^{1} \sin^{-1}(u)\cdot 2u\,du = 2\int_{0}^{1} u\,\sin^{-1}u\,du.$$

Now, apply integration by parts:

• Let $v=\sin^{-1}u$ so that $\frac{dv}{du}=\frac{1}{\sqrt{1-u^2}}$.
• Let $dw= u\,du$ so that $w=\frac{u^2}{2}$.

Thus,

$$\int u\,\sin^{-1}u\,du = \frac{u^2}{2}\sin^{-1}u - \int \frac{u^2}{2}\cdot \frac{1}{\sqrt{1-u^2}}\,du.$$

Multiplying by 2, we have:

$$2\int_0^1 u\,\sin^{-1}u\,du = \left[\ u^2\sin^{-1}u\ \right]_0^1 - \int_0^1 \frac{u^2}{\sqrt{1-u^2}}\,du.$$

At $u=1$, $\sin^{-1}1=\frac{\pi}{2}$ and $1^2=1$. At $u=0$, the term vanishes. Hence,

$$2\int_0^1 u\,\sin^{-1}u\,du = \frac{\pi}{2} - \int_0^1 \frac{u^2}{\sqrt{1-u^2}}\,du.$$

Now, evaluate

$$K = \int_0^1 \frac{u^2}{\sqrt{1-u^2}}\,du.$$

Substitute $u=\sin\theta$ so that $du=\cos\theta\,d\theta$ and $\sqrt{1-u^2}=\cos\theta$. The limits transform to: when $u=0$, $\theta=0$; when $u=1$, $\theta=\pi/2$. Then,

$$K = \int_{0}^{\pi/2} \frac{\sin^2\theta}{\cos\theta}\cdot \cos\theta\,d\theta = \int_{0}^{\pi/2} \sin^2\theta\,d\theta.$$

A standard integral gives:

$$\int_{0}^{\pi/2} \sin^2\theta\,d\theta = \frac{\pi}{4}.$$

Thus,

$$2\int_0^1 u\,\sin^{-1}u\,du = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.$$

This means $J = \frac{\pi}{4}$, and hence

$$I = \frac{\pi}{4}.$$