∫cos^(−1) (sinx)dx=?


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Question Number 45670 by arvinddayama01@gmail.com last updated on 15/Oct/18

$\int {c o s}^{- 1} (s i n x) d x = ?$
Commented by maxmathsup by imad last updated on 15/Oct/18

$l e t I = \int a r c c o s (s i n x) d x c h a n g e m e n t a r c o s (s i n x) = t \Rightarrow s i n x = c o s t$ $\Rightarrow x = a r c s i n (c o s t) \Rightarrow d x = - s i n t \frac{1}{\sqrt{1 - {c o s}^{2} t}} d t = - d t \Rightarrow$ $I = \int t (- d t) = - \frac{t^{2}}{2} + c = - \frac{1}{2} {(a r c o s (s i n x))}^{2} + c .$
	Answered by ARVIND DAYAMA last updated on 15/Oct/18
	$∵cos^(−1) (sinx)=t −(1/(√(1−sin^2 x))).cosxdx=dt dx=−dt −∫tdt −(1/2)t^2 +C −(1/2){cos^(−1) (sinx)}^2 +C$
	$∵ {c o s}^{- 1} (s i n x) = t$ $- \frac{1}{\sqrt{1 - {s i n}^{2} x}} . c o s x d x = d t$ $d x = - d t$ $- \int t d t$ $- \frac{1}{2} t^{2} + C$ $- \frac{1}{2} {{c o s}^{- 1} (s i n x)}^{2} + C$
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