Question

Integrate the following
i) $\log x$
ii) ${\cos }^{-1}\left(\sqrt{x}\right)$

Answer 1

(i) ${\int }^{}\log xdx$

$={\int }^{}1.\log xdx$

$=x\log x-{\int }^{}\frac{1}{x}.xdx$

$=x\log x-x+c$

(ii) ${\int }^{}{\cos }^{-1}\sqrt{x}dx$

putting $x=t^2$

$\Rightarrow dx=2tdt$

$=2{\int }^{}t{\cos }^{-1}tdt$

$=2\left[\frac{t^2}{2}{\cos }^{-1}t-{\int }^{}\frac{-1}{\sqrt{1-t^2}}\times \frac{t^2}{2}dt\right]$

$=2\left[\frac{t^2}{2}{\cos }^{-1}t-\frac{1}{2}{\int }^{}\frac{1-t^2+1}{\sqrt{1-t^2}}dt\right]$

$=2\left[\frac{t^2}{2}{\cos }^{-1}t-\frac{1}{2}{\int }^{}\left(\sqrt{1-t^2}+\frac{1}{\sqrt{1-t^2}}\right)dt\right]$

$=t^2{\cos }^{-1}t-{\int }^{}\sqrt{1-t^2}dt-{\int }^{}\frac{1}{\sqrt{1-t^2}}dt$

$=t^2{\cos }^{-1}t-\left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t\right]-{\sin }^{-1}t+c$

$=t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}-\frac{3}{2}{\sin }^{-1}t+c$

$=x{\cos }^{-1}\sqrt{x}-\frac{\sqrt{x}}{2}\sqrt{1-x}-\frac{3}{2}\sin {}^{-1}\sqrt{x}+c$

$=x{\cos }^{-1}\sqrt{x}-\frac{\sqrt{x-x^2}}{2}-\frac{3}{2}\sin {}^{-1}\sqrt{x}+c$