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Integral of arctan((tanx−2cotx)/3) | JEE

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

Question

Let

\int \tan^{-1}\!\left(\dfrac{\tan x}{2}\right)dx=\alpha

. Then for

0<x<\dfrac{\pi}{2}

, the value of

\displaystyle\int \tan^{-1}\!\left(\dfrac{\tan x-2\cot x}{3}\right)dx

equals

\alpha+\dfrac{\pi x}{2}+\dfrac{x^{2}}{2}+c

\alpha-\dfrac{\pi x}{2}+\dfrac{x^{2}}{2}+c

correct

-\alpha-\dfrac{\pi x}{2}+\dfrac{x^{2}}{2}+c

-\alpha+\dfrac{\pi x}{2}-\dfrac{x^{2}}{2}+c

Solution

Step 1: Recognise the arctangent difference law. Since

\tan x\cdot\cot x=1

, the constant

3

in the denominator can be read as

1+2\tan x\cot x

\dfrac{\tan x-2\cot x}{3}=\dfrac{\tan x-2\cot x}{1+(\tan x)(2\cot x)}

This is precisely the expansion of a difference of arctangents:

\tan^{-1}\!\left(\dfrac{\tan x-2\cot x}{3}\right)=\tan^{-1}(\tan x)-\tan^{-1}(2\cot x)

Step 2: Evaluate each inverse function on

\left(0,\dfrac{\pi}{2}\right)

. Here

\tan^{-1}(\tan x)=x

. For the second term,

2\cot x=\dfrac{2}{\tan x}

, so by the complementary identity

\tan^{-1}(2\cot x)=\tan^{-1}\!\left(\dfrac{2}{\tan x}\right)=\dfrac{\pi}{2}-\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)

Step 3: Substitute these back into the integrand:

\tan^{-1}\!\left(\dfrac{\tan x-2\cot x}{3}\right)=x-\left[\dfrac{\pi}{2}-\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)\right]=x-\dfrac{\pi}{2}+\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)

Step 4: Integrate term by term, using

\displaystyle\int\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)dx=\alpha

as given:

I=\int\left[x-\dfrac{\pi}{2}+\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)\right]dx=\dfrac{x^{2}}{2}-\dfrac{\pi x}{2}+\alpha+c

Correct answer: (2)

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