Question

$\int\limits_0^{\infty} \frac{x^3 d x}{\left(1+x^2\right)^{9 / 2}}$ is equal to

• A. $\frac{2}{35}$
• B. $\frac{4}{35}$
• C. $\frac{6}{35}$
• D. $\frac{3}{35}$

Answer 1

Answer: (A)

Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
$\Rightarrow \int\limits_0^{\infty} \frac{x^3}{\left(1+x^2\right)^{9 / 2}} d x=\int\limits_0^{\pi / 2} \frac{\tan ^3 \theta}{\sec ^9 \theta} \cdot \sec ^2 \theta d \theta$
$=\int\limits_0^{\pi / 2} \sin ^3 \theta \cos ^4 \theta d \theta=\frac{2.3 .1}{7.5 .3 .1}=\frac{2}{35}$