Indefinite Integration
Integration by Substitution
Grade None
Question:
<p>[JEE Main 2022] \(\displaystyle\int\frac{dx}{1+3\sin^2 x+8\cos^2 x}\) equals (where \(C\) is constant)</p>
<li>\(\dfrac{1}{6}\tan^{-1}\!\dfrac{2\tan x}{3}+C\)</li>
<li>\(\dfrac{1}{3}\tan^{-1}(2\tan x)+C\)</li>
<li>\(\dfrac{1}{6}\tan^{-1}\!\dfrac{3\tan x}{2}+C\)</li>
<li>\(\dfrac12\tan^{-1}\!\dfrac{\tan x}{3}+C\)</li>
Step-by-Step Solution
Key Concept: Write sin^2x=tan^2x/(1+tan^2x), cos^2x=1/(1+tan^2x). Multiply through by sec^2x. Substitute t=tanx.
<p>Divide numerator and denominator by $\cos^2 x$:</p>
<p>$$\int\frac{\sec^2 x}{\sec^2 x+3\tan^2 x+8}\,dx = \int\frac{\sec^2 x}{(1+\tan^2 x)+3\tan^2 x+8}\,dx$$</p>
<p>$$= \int\frac{\sec^2 x}{4\tan^2 x+9}\,dx$$</p>
<p>Let $t=\tan x,\;dt=\sec^2 x\,dx$:</p>
<p>$$= \int\frac{dt}{4t^2+9} = \frac{1}{4}\int\frac{dt}{t^2+(3/2)^2} = \frac{1}{4}\cdot\frac{1}{3/2}\tan^{-1}\!\frac{t}{3/2}+C = \frac{1}{6}\tan^{-1}\!\frac{2\tan x}{3}+C$$</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A