Question

Use integration by parts to evaluate
${\int }_1^2\cos h^{-1}xdx$

Answer 1

Formula: $\cos h^{-1}x=\ln \left(x+\sqrt{x^2-1}\right)$

and $\frac{d}{dx}\cos h^{-1}x=\frac{1}{\sqrt{x^2-1}}$

Using the rule, $\int udv=uv-\int vdu$

THerefore, $\int \cos h^{-1}x=\cos h^{-1}x\times \int 1.dx-\int \frac{x}{\sqrt{x^2-1}}dx$

$\Rightarrow x.\cos h^{-1}x-\sqrt{x^2-1}$

Substituting the limits of integration, we get

$\Rightarrow |x.\cos h^{-1}x-\sqrt{x^2-1}{|}_1^2$

$\Rightarrow (2\cos h^{-1}2-1\cos h^{-1}1)-(\sqrt{2^2-1}-\sqrt{1^2-1})$

$\Rightarrow \left(2\ln \right(2+\sqrt{3})-\ln 1)-(\sqrt{3}-0)$

$\Rightarrow 2\ln (2+\sqrt{3})-\sqrt{3}$

Hence, ${\int }_1^2\cos h^{-1}x=2\ln (2+\sqrt{3})-\sqrt{3}$