Question

Differentiate $\sin h^{-1}x$ with respect to $x$.
By writing $\sin h^{-1}x$ as $1\times \sin h^{-1}x$, use integration by parts to find ${\int }_1^2\sin h^{-1}dx$

Answer 1

$\frac{d}{dx}\left(sin{h}^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}$

Now ${\int }_1^{2}sin{h}^{-1}xdx={\int }_1^{2}1\times sin{h}^{-1}xdx$

First let us evaluate $\int 1\times sin{h}^{-1}xdx$

Let $u=sin{h}^{-1}x\Rightarrow du=\frac{1}{\sqrt{1+x^2}}dx$ and $dv=1dx\Rightarrow v=x$

$\therefore \int 1\times sin{h}^{-1}xdx=xsin{h}^{-1}x-\int x\cdot \frac{1}{\sqrt{1+x^2}}dx$

$=xsin{h}^{-1}x-\int \frac{x}{\sqrt{1+x^2}}dx$

Comsider $\int \frac{x}{\sqrt{1+x^2}}dx$

Let $w=1+x^2\Rightarrow dw=2xdx$

$\Rightarrow \int \frac{x}{\sqrt{1+x^2}}dx=\int \frac{1}{2\sqrt{w}}dw=\sqrt{w}=\sqrt{1+x^2}$

$\therefore \int 1\times sin{h}^{-1}xdx=xsin{h}^{-1}x-\int \frac{x}{\sqrt{1+x^2}}dx=xsin{h}^{-1}x-\sqrt{1+x^2}$

Now ${\int }_1^{2}1\times sin{h}^{-1}xdx={[xsin{h}^{-1}x-\sqrt{1+x^2}]}_1^2$

$=2sin{h}^{-1}2-\sqrt{5}-sin{h}^{-1}1+\sqrt{2}$

$=2ln(\sqrt{5}+2)-\sqrt{5}-ln(\sqrt{2}+1)+\sqrt{2}$