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Integral of cotx/√(5+9cot²x) | JEE

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

Question

\displaystyle\int \dfrac{\cot x}{\sqrt{5+9\cot^{2}x}}\,dx

equals

\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{2\sin x}{3}\right)+C

correct

\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{3\sin x}{2}\right)+C

\dfrac{1}{3}\sin^{-1}\!\left(\dfrac{3\sin x}{2}\right)+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

. Since

\cot x=\dfrac{\cos x}{\sin x}

and

5+9\cot^{2}x=\dfrac{5\sin^{2}x+9\cos^{2}x}{\sin^{2}x}

\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^{2}x=1-\sin^{2}x

5\sin^{2}x+9\cos^{2}x=5\sin^{2}x+9(1-\sin^{2}x)=9-4\sin^{2}x

\Rightarrow I=\int\dfrac{\cos x}{\sqrt{9-4\sin^{2}x}}\,dx

Step 3: Substitute

t=\sin x\Rightarrow dt=\cos x\,dx

I=\int\dfrac{dt}{\sqrt{9-4t^{2}}}

Step 4: Apply

\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=2

I=\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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