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
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