Evaluate the following integrals:


Evaluate: $\int \cos ^{-1}(\sin x) d x$


Given, $\int \cos ^{-1}(\sin x) d x$

Let us consider, $\int \cos ^{-1} \mathrm{dx}$

We know that, $\int f(x) \cdot g(x) d x=f(x) \int g(x) d x-\int\left[f^{\prime}(x) \int g(x)\right] d x$

By comparison, $f(x)=\cos ^{-1} x ; g(x)=1$

$=\cos ^{-1} x x \int 1 d x-\int-\frac{1}{\sqrt{1-x^{2}}} \cdot x d x$

$=x \cos ^{-1} x-\frac{1}{2} \int \frac{1}{\sqrt{1-x^{2}}}(-2 x) d x$

$=x \cos ^{-1} x-\frac{1}{2} \int\left(1-x^{2}\right)^{-\frac{1}{2}}(-2 x) d x$

$=x \cos ^{-1} x-\frac{1}{2} \frac{\left(1-x^{2}\right)^{-\frac{1}{2}+1}}{\frac{1}{2}+1}+c\left(\right.$ since, $\left.\int\left[f(x)^{n} \cdot f^{I}(x)\right] d x=\frac{f(x)^{n+1}}{n+1}\right)$

$=x \cos ^{-1} x-\left(1-x^{2}\right)^{1 / 2}+c$

$=x \cos ^{-1} x-\sqrt{1-x^{2}}+c$

Therefore, $\int \cos ^{-1} x d x=x \cos ^{-1} x-\sqrt{1-x^{2}}+c$

Replace ' $x$ ' with ' $\sin x$ ' :-

$\int \cos ^{-1}(\sin x) d x=\sin x \cdot \cos ^{-1}(\sin x)-\sqrt{1-(\sin x)^{2}}+c$

$=\sin x \cdot \cos ^{-1} x(\sin x)-\cos x+c$

Leave a comment