Mathematics - Definite Integration Question with Solution | TestHub

To evaluate the integral ${\int }_0^{\pi }{\cos }^{-1}(\sin x)dx$, we can use the property of definite integrals. Since ${\cos }^{-1}(\sin x)$ is symmetric about $x=\frac{\pi }{2}$ over the interval $[0,\pi ]$, we can write:

$${\int }_0^{\pi }{\cos }^{-1}(\sin x)dx=2\left[{\int }_0^{\pi /2}{\cos }^{-1}(\sin x)dx\right]$$

Using the identity ${\cos }^{-1}(y)=\frac{\pi }{2}-{\sin }^{-1}(y)$, we have:

$$2\left[{\int }_0^{\pi /2}\left(\frac{\pi }{2}-{\sin }^{-1}(\sin x)\right)dx\right]$$

For $x\in [0,\frac{\pi }{2}]$, ${\sin }^{-1}(\sin x)=x$. Therefore, the integral becomes:

$$2\left[{\int }_0^{\pi /2}\left(\frac{\pi }{2}-x\right)dx\right]$$

Now, we evaluate the integral:

$$2{\left[\frac{\pi }{2}x-\frac{x^2}{2}\right]}_0^{\pi /2}$$

$$2\left[\left(\frac{\pi }{2}\cdot \frac{\pi }{2}-\frac{(\pi /2{)}^2}{2}\right)-(0-0)\right]$$

$$2\left[\frac{{\pi }^2}{4}-\frac{{\pi }^2/4}{2}\right]$$

$$2\left[\frac{{\pi }^2}{4}-\frac{{\pi }^2}{8}\right]$$

$$2\left[\frac{2{\pi }^2-{\pi }^2}{8}\right]$$

$$2\left[\frac{{\pi }^2}{8}\right]$$

$$\frac{{\pi }^2}{4}$$

Thus, the value of the integral is $\frac{{\pi }^2}{4}$.