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Integral of cotx/√(5+9cot²x) | JEE

JEE Maths question with a full step-by-step solution.

Question
∫cot⁡x5+9cot⁡2x dx\displaystyle\int \dfrac{\cot x}{\sqrt{5+9\cot^{2}x}}\,dx equals
A12sin⁡−1 ⁣(2sin⁡x3)+C\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{2\sin x}{3}\right)+Ccorrect
B12sin⁡−1 ⁣(3sin⁡x2)+C\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{3\sin x}{2}\right)+C
C13sin⁡−1 ⁣(3sin⁡x2)+C\dfrac{1}{3}\sin^{-1}\!\left(\dfrac{3\sin x}{2}\right)+C
D13sin⁡−1 ⁣(2sin⁡x3)+C\dfrac{1}{3}\sin^{-1}\!\left(\dfrac{2\sin x}{3}\right)+C
Solution
Step 1: Clear the cotangents by multiplying numerator and denominator through by sin⁡x\sin x. Since cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x} and 5+9cot⁡2x=5sin⁡2x+9cos⁡2xsin⁡2x5+9\cot^{2}x=\dfrac{5\sin^{2}x+9\cos^{2}x}{\sin^{2}x},
cot⁡x5+9cot⁡2x=cos⁡x5sin⁡2x+9cos⁡2x\dfrac{\cot x}{\sqrt{5+9\cot^{2}x}}=\dfrac{\cos x}{\sqrt{5\sin^{2}x+9\cos^{2}x}}
 Step 2: Simplify the radicand with cos⁡2x=1−sin⁡2x\cos^{2}x=1-\sin^{2}x:
5sin⁡2x+9cos⁡2x=5sin⁡2x+9(1−sin⁡2x)=9−4sin⁡2x5\sin^{2}x+9\cos^{2}x=5\sin^{2}x+9(1-\sin^{2}x)=9-4\sin^{2}x
⇒I=∫cos⁡x9−4sin⁡2x dx\Rightarrow I=\int\dfrac{\cos x}{\sqrt{9-4\sin^{2}x}}\,dx
 Step 3: Substitute t=sin⁡x⇒dt=cos⁡x dxt=\sin x\Rightarrow dt=\cos x\,dx:
I=∫dt9−4t2I=\int\dfrac{dt}{\sqrt{9-4t^{2}}}
 Step 4: Apply ∫dta2−b2t2=1bsin⁡−1 ⁣(bta)\displaystyle\int\dfrac{dt}{\sqrt{a^{2}-b^{2}t^{2}}}=\dfrac{1}{b}\sin^{-1}\!\left(\dfrac{bt}{a}\right) with a=3, b=2a=3,\ b=2:
I=12sin⁡−1 ⁣(2t3)+C=12sin⁡−1 ⁣(2sin⁡x3)+CI=\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{2t}{3}\right)+C=\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{2\sin x}{3}\right)+C
 Correct answer: (1)
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