Indefinite Integration
Integration of Trigonometric Functions
Grade 12
Question:
<p>[JEE Main 2019] \(\displaystyle\int\frac{2\sin x}{2+\sin 2x}\,dx\) equals (where \(C\) is a constant)</p>
<li>\(-\ln|1+\tan x|+\dfrac{1}{2}\ln(1+\tan^2 x)+C\)</li>
<li>\(\ln|1+\tan x|-\dfrac{1}{2}\ln(1+\tan^2 x)+C\)</li>
<li>\(\dfrac{1}{2}\tan^{-1}(\tan x+1)+C\)</li>
<li>\(-2\tan^{-1}(1+\tan x)+C\)</li>
Step-by-Step Solution
Key Concept: Write sin2x=2sinx cosx, so 2+sin2x=2+2sinx cosx. Factor and substitute t=tanx.
<p>$2+\sin 2x = 2+2\sin x\cos x = 2\cos^2 x(1+\tan x)+2\sin^2 x\cdots$ Hmm.</p>
<p>Better: divide top and bottom by $\cos^2 x$:</p>
<p>$$\frac{2\sin x/\cos^2 x}{2/\cos^2 x+2\sin x/\cos x} = \frac{2\tan x\sec x}{2\sec^2 x+2\tan x\sec x}\cdots$$</p>
<p>Use $2+\sin 2x=(1+\sin x+\cos x)^2-(\sin x-\cos x)^2\cdots$ or just note:</p>
<p>$$\int\frac{2\sin x}{2+2\sin x\cos x}\,dx = \int\frac{\sin x}{1+\sin x\cos x}\,dx$$</p>
<p>Divide by $\cos^2 x$: $=\int\frac{\tan x\sec x}{\sec^2 x+\tan x}\,dt$ with $t=\tan x$:</p>
<p>$$=\int\frac{t}{(1+t^2)+t}\,dt = \int\frac{t}{t^2+t+1}\,dt$$ which gives the log form. Answer: <strong>(A)</strong></p>
Correct Answer: A