Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Integrate the function\[\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\]

$\begin{array}{1 1} -[\cos^{-1}x\sqrt{1-x^2}+x]+c \\ [\cos^{-1}x\sqrt{1+x^2}+x]+c \\-[\cos^{-1}x\sqrt{1-x^2}-x]+c \\ [\cos^{-1}x\sqrt{1+x^2}-x]+c \end{array} $

Can you answer this question?

1 Answer

0 votes
  • (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)Method of substitution $\int f(x)dx,$ if we substitute f(x) as t,then $f'(x)dx=dt$ hence the integral function becomes $\int t.dt$
  • (iii)$\frac{d}{dx}\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}.$
Given $ I=\frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx.$
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts
Let $u=\cos{-1}x.$
Differentiating with respect to x we get
Let $dv=\frac{x}{\sqrt{1-x^2}}dx.$
This can be integrated by the method of substitution
Let $(1-x^2)=t.$
On differentiating we get
$\Rightarrow xdx=\frac{-dt}{2}.$
Now substituting t and dt
$\int dv=\frac{-1}{2}\int\frac{dt}{\sqrt t}.$
On integrating we get
v=$\frac{-1}{2}(\sqrt t).$
Substituting for t we get
Now substituting for u,v,du and dv we get,
On canceling the common terms,


answered Feb 10, 2013 by sreemathi.v
edited Jul 21, 2013 by balaji.thirumalai
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App