If $f(x)=\int\limits_0^{\sin x} \cos ^{-1} t d t+\int\limits_0^{\cos x} \sin ^{-1} t d t$, $0<x<\frac{\pi}{2}$, then $f(\pi / 4)$ is equal to

$1+\frac{\pi}{2 \sqrt{2}}$

We have,

$f(x) =\int\limits_0^{\sin x} \cos ^{-1} t d t+\int\limits_0^{\cos x} \sin ^{-1} t d t$

$\Rightarrow f(x) =\int\limits_0^{\sin x}\left(\frac{\pi}{2}-\sin ^{-1} t\right) d t+\int\limits_0^{\cos x}\left(\frac{\pi}{2}-\cos ^{-1} t\right) d t$

$\Rightarrow f(x)=\frac{\pi}{2} \sin x+\frac{\pi}{2} \cos x-\int\limits_0^{\sin x} \sin ^{-1} t d t-\int\limits_0^{\cos x} \cos ^{-1} t d t$

$\Rightarrow f(x)=\frac{\pi}{2}(\sin x+\cos x)-\left[\int\limits_0^{\sin x} \sin ^{-1} t d t+\int\limits_0^{\cos x} \cos ^{-1} t d t\right]$

Now, $\int\limits_0^{\sin x} \sin ^{-1} t d t=\int\limits_0^x \theta \cos \theta d \theta$, where $t=\sin \theta$

$\Rightarrow \int\limits_0^{\sin x} \sin ^{-1} t d t=[\theta \sin \theta+\cos \theta]_0^x=x \sin x+\cos x-1$

and,

$\int\limits_0^{\cos x} \cos ^{-1} t d t =-\int\limits_0^x \alpha \sin \alpha d \alpha$, where $t=\cos \alpha$

$\Rightarrow \int\limits_0^{\cos x} \cos ^{-1} t d t =-[-\alpha \cos \alpha+\sin \alpha]_0^x$

$\Rightarrow \int\limits_0^{\cos x} \cos ^{-1} t d t =x \cos x-\sin x$

∴  $f(x)=\frac{\pi}{2}(\sin x+\cos x) -[x \sin x+\cos x-1+x \cos x-\sin x]$

$\Rightarrow f\left(\frac{\pi}{4}\right)=\frac{\pi}{2}\left(\frac{2}{\sqrt{2}}\right)-\left[\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1+\frac{\pi}{4 \sqrt{2}}-\frac{1}{\sqrt{2}}\right]$

$\Rightarrow f\left(\frac{\pi}{4}\right)=\frac{\pi}{\sqrt{2}}-\frac{\pi}{2 \sqrt{2}}+1=\frac{\pi}{2 \sqrt{2}}+1$