# 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

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

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