Question

The value of $\int \frac{dx}{{\left(1-x^2\right)}^{3/2}}$ equals

• A. $x\sqrt{1-x^2}+c$
• B. $\frac{x}{2\sqrt{1-x^2}}+c$
• C. $\frac{x}{\sqrt{1-x^2}}+c$
• D. $\frac{2x}{\sqrt{1-x^2}}+c$

Answer 1

Answer: (C)

$\frac{x}{\sqrt{1-x^2}}+c$

$\int \frac{dx}{{(1-x^2)}^{\frac{3}{2}}}$

Take $x=\sin \theta \Rightarrow \cos \theta =\sqrt{1-x^2}$

$\Rightarrow dx=\cos \theta d\theta$

$\int \frac{dx}{{(1-x^2)}^{\frac{3}{2}}}=\int \frac{\cos \theta d\theta }{{(1-{\sin }^2\theta )}^{\frac{3}{2}}}$

$=\int \frac{\cos \theta d\theta }{{\left({\cos }^2\theta \right)}^{\frac{3}{2}}}$

$=\int \frac{d\theta }{{\cos }^2\theta }$

$=\int {\sec }^2\theta d\theta$

$=\tan \theta +c$ where $c$ is the constant of integration

$=\frac{\sin \theta }{\cos \theta }$

$=\frac{x}{\sqrt{1-x^2}}$ since $x=\sin \theta$ and $\sqrt{1-x^2}=\cos \theta$