Question

$\int \frac{\mathrm{dx}}{4+5{\sin }^{2}x}\mathrm{dx}$ is equal to.

• A.
  $\frac{1}{6}{\tan }^{-1}\left(\frac{3\mathrm{cotx}}{2}\right)+C$
• B.
  $\frac{1}{6}{\tan }^{-1}\left(\frac{3\mathrm{tanx}}{2}\right)+C$
• C.
  $\frac{1}{6}{\cot }^{-1}\left(\frac{3\mathrm{cotx}}{2}\right)+C$
• D.
  $\frac{1}{6}{\cot }^{-1}\left(\frac{3\mathrm{tanx}}{2}\right)+C$

Answer 1

The correct option is B

$\frac{1}{6}{\tan }^{-1}\left(\frac{3\mathrm{tanx}}{2}\right)+C$
These types of integrals of the form $\int \frac{dx}{a+b{\sin }^{2}x}$ or $\int \frac{\mathrm{dx}}{a+{\mathrm{bcos}}^{2}x}$
are solved by multiplying ${\sec }^{2}x$ in both numerator and denominator of the integrand and then substituting $t=\mathrm{tanx}$.
So, for our integral multiplying ${\sec }^{2}x$ in both numerator and denominator we get,
$I=\int \frac{{\sec }^{2}x}{(4+5{\sin }^{2}x){\sec }^{2}x}\mathrm{dx}$
$\Rightarrow I=\int \frac{{\sec }^{2}x\mathrm{dx}}{4{\sec }^{2}x+5{\sin }^{2}x{\sec }^{2}x}$
$\Rightarrow I=\int \frac{{\sec }^{2}x\mathrm{dx}}{4(1+{\tan }^{2}x)+5{\tan }^{2}x}\Rightarrow I=\int \frac{{\sec }^{2}x\mathrm{dx}}{4+9{\tan }^{2}x}$
Now, if we assume $t=\mathrm{tanx}$
we get $\mathrm{dt}={\sec }^2\mathrm{xdx}$
So, our integral becomes $I=\int \frac{\mathrm{dt}}{4+9t^2}\Rightarrow I=\frac{1}{9}\int \frac{\mathrm{dt}}{{\left(\frac{2}{3}\right)}^2+t^2}\Rightarrow I=\frac{1}{9}\times \frac{3}{2}{\tan }^{-1}\left(\frac{3t}{2}\right)+C\Rightarrow I=\frac{1}{6}{\tan }^{-1}\left(\frac{3\mathrm{tanx}}{2}\right)+C$