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Find the integrals of the functions\[\sin^{-1}(\cos x)\]

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

1 Answer

  • $(i)\;\int\frac{1}{\sqrt{1-x^2}dx}=\sin^{-1}x+c.$
  • $(ii)\;$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,
$-\sin xdx=dt.$
$\sin xdx=-dt.\Rightarrow dx=\frac{-dt}{\sin x}.$
Now let $\sin^{-1}t=u.$
On differentiating we get,
On substituting we get,
$I=\int\sin^{-1}(\cos x)dx=-\int u.du.$
On integrating we get,
Substituting for u we get,
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