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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\sin^{-1}(\cos x)\]

$\begin{array}{1 1}\frac{-\sin^{-1}(\cos x)}{2}+c \\\frac{-\cos^{-1}(\cos x)}{2}+c \\ \frac{-\sin^{-1}(\sin x)}{2}+c \\ -\sin^{-1}(\cos x)+c\end{array}$

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1 Answer

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Toolbox:
  • $(i)\;\int\frac{1}{\sqrt{1-x^2}dx}=\sin^{-1}x+c.$
  • $(ii)\;$Method of substitution:
  • If f(x)=t,then f,(x)dx=dt.
  • Thus $\int f(x)dx=\int t.dt$.
Given $I=\int\sin^{-1}(\cos x)dx.$
 
Let $\cos x=t.$
 
$\sin x=\sqrt{1-t^2}$.($1-\cos^2x=\sin^2x.)$
 
On integrating we get,
 
$-\sin xdx=dt.$
 
$\sin xdx=-dt.\Rightarrow dx=\frac{-dt}{\sin x}.$
 
$\qquad\qquad=\frac{-dt}{\sqrt{1-t^2}}.$
 
Now let $\sin^{-1}t=u.$
 
On differentiating we get,
 
$\frac{1}{\sqrt{1-t^2}}dt=du.$
 
On substituting we get,
 
$I=\int\sin^{-1}(\cos x)dx=-\int u.du.$
 
On integrating we get,
 
$\;\;\;=\frac{-u^2}{2}+c.$
 
Substituting for u we get,
 
$\;\;\;=\frac{-\sin^2t}{2}+c.$
 
Substituting for t we get,
 
$\int\sin^{-1}(\cos x)dx=\frac{-\sin^{-1}(\cos x)}{2}+c.$

 

answered Feb 3, 2013 by sreemathi.v
 
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