Question
Integrate the function $\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}$ with respect to $x$.
✓
Answer
Let
$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x$
$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right) \sqrt{1-x^2}} d x$
$\text { Putting } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$
$I=\int \frac{t d t}{\left(1-\sin ^2 t\right)}=\int \frac{t d t}{\cos ^2 t}=\int t \sec ^2 t d t$
$I=\int_{I}^{t II} t \sec ^2 t d t$
$\text { (using ILATE) }$
$I=t \cdot \tan t-\int 1 \cdot \tan t d t$
$=t \tan t-\log \sec t+C$
$\therefore I=\sin ^{-1} x \tan \left(\sin ^{-1} x\right)-\log \sec \left(\sin ^{-1} x\right)+C$
$I=\sin ^{-1} x \tan \left(\tan ^{-1} \frac{x}{\sqrt{1-x^2}}\right)-$
$\log \sec \left(\sec ^{-1} \frac{1}{\sqrt{1-x^2}}\right)+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-\log \frac{1}{\sqrt{1-x^2}}+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-\log \left(1-x^2\right)^{-\frac{1}{2}}+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{1}{2} \log \left(1-x^2\right)+C$
$\text { }$
$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x$
$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right) \sqrt{1-x^2}} d x$
$\text { Putting } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$
$I=\int \frac{t d t}{\left(1-\sin ^2 t\right)}=\int \frac{t d t}{\cos ^2 t}=\int t \sec ^2 t d t$
$I=\int_{I}^{t II} t \sec ^2 t d t$
$\text { (using ILATE) }$
$I=t \cdot \tan t-\int 1 \cdot \tan t d t$
$=t \tan t-\log \sec t+C$
$\therefore I=\sin ^{-1} x \tan \left(\sin ^{-1} x\right)-\log \sec \left(\sin ^{-1} x\right)+C$
$I=\sin ^{-1} x \tan \left(\tan ^{-1} \frac{x}{\sqrt{1-x^2}}\right)-$
$\log \sec \left(\sec ^{-1} \frac{1}{\sqrt{1-x^2}}\right)+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-\log \frac{1}{\sqrt{1-x^2}}+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-\log \left(1-x^2\right)^{-\frac{1}{2}}+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{1}{2} \log \left(1-x^2\right)+C$
$\text { }$
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