Mathematics

# $\displaystyle \int { \dfrac { 1 }{ 2 } \cos ^{ -1 }{ x } dx }$

##### SOLUTION
Let, $\dfrac{1}{2}cos^{-1}x=t$

$\implies x=cos2t$

$\implies dx=-2sin2tdt$

$\implies \int -2t.sin2t.dt$

$\implies -2\int t.sin2t.dt$

Solving this, By parts, we get

$\implies -2[\dfrac{-t.cos2t}{2}+\dfrac{sin2t}{4}]+C$

$\implies t.cos2t+\dfrac{sin2t}{2}+C$

$\implies \dfrac{1}{2}cos^{-1}x.cos2(\dfrac{1}{2}cos^{-1}x)+\dfrac{sin2(\dfrac{1}{2}cos^{-1}x)}{2}+C$

$\implies \dfrac{1}{2}(cos^{-1}x)^2+\dfrac{sin(cos^{-1}x)}{2}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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