Mathematics

# Integrate the followingi) $\log{x}$ii) ${\cos}^{-1}(\sqrt{x})$

##### SOLUTION
(i)  $\int_{}^{} {\log xdx}$
$= \int_{}^{} {1.\log xdx}$
$= x\log x - \int_{}^{} {\cfrac{1}{x}.xdx}$
$= x\log x - x + c$
(ii)  $\int_{}^{} {{{\cos }^{ - 1}}\sqrt x dx}$
putting $x = {t^2}$
$\Rightarrow dx = 2tdt$
$= 2\int_{}^{} {t{{\cos }^{ - 1}}tdt}$
$= 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \int_{}^{} {\cfrac{{ - 1}}{{\sqrt {1 - {t^2}} }} \times \cfrac{{{t^2}}}{2}dt} } \right]$
$= 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \cfrac{1}{2}\int_{}^{} {\cfrac{{1 - {t^2} + 1}}{{\sqrt {1 - {t^2}} }}dt} } \right]$
$= 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \cfrac{1}{2}\int_{}^{} {\left( {\sqrt {1 - {t^2}} + \cfrac{1}{{\sqrt {1 - {t^2}} }}} \right)dt} } \right]$
$= {t^2}{\cos ^{ - 1}}t - \int_{}^{} {\sqrt {1 - {t^2}} dt - \int_{}^{} {\cfrac{1}{{\sqrt {1 - {t^2}} }}dt} }$
$= {t^2}{\cos ^{ - 1}}t - \left[ {\cfrac{t}{2}\sqrt {1 - {t^2}} + \cfrac{1}{2}{{\sin }^{ - 1}}t} \right] - {\sin ^{ - 1}}t + c$
$= {t^2}{\cos ^{ - 1}}t - \cfrac{t}{2}\sqrt {1 - {t^2}} - \cfrac{3}{2}{\sin ^{ - 1}}t + c$
$= x{\cos ^{ - 1}}\sqrt x - \cfrac{{\sqrt x }}{2}\sqrt {1 - x} - \cfrac{3}{2}{\sin {^{ - 1}\sqrt x }} + c$
$= x{\cos ^{ - 1}}\sqrt x - \cfrac{{\sqrt {x - {x^2}} }}{2} - \cfrac{3}{2}{\sin {^{ - 1}\sqrt x }} + c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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