Indefinite IntegrationmediumFree

∫eˣ(2tanx/(1+tanx)+cot²(x+π/4)) dx | JEE

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

Question

\displaystyle\int e^{x}\left(\dfrac{2\tan x}{1+\tan x}+\cot^{2}\!\left(x+\dfrac{\pi}{4}\right)\right)dx

equals

e^{x}\tan\!\left(\dfrac{\pi}{4}-x\right)+C

e^{x}\tan\!\left(x-\dfrac{\pi}{4}\right)+C

correct

e^{x}\tan\!\left(\dfrac{3\pi}{4}-x\right)+C

e^{x}\tan\!\left(x-\dfrac{3\pi}{4}\right)+C

Solution

Step 1: Convert the cotangent term. Since

\cot\!\left(x+\tfrac{\pi}{4}\right)=-\tan\!\left(x-\tfrac{\pi}{4}\right)

, squaring gives

\cot^{2}\!\left(x+\tfrac{\pi}{4}\right)=\tan^{2}\!\left(x-\tfrac{\pi}{4}\right)=\sec^{2}\!\left(x-\tfrac{\pi}{4}\right)-1

Step 2: Use the

-1

to simplify the first term:

\dfrac{2\tan x}{1+\tan x}-1=\dfrac{2\tan x-(1+\tan x)}{1+\tan x}=\dfrac{\tan x-1}{1+\tan x}=\tan\!\left(x-\dfrac{\pi}{4}\right)

Step 3: Combine the bracket:

\dfrac{2\tan x}{1+\tan x}+\cot^{2}\!\left(x+\tfrac{\pi}{4}\right)=\left[\dfrac{2\tan x}{1+\tan x}-1\right]+\sec^{2}\!\left(x-\tfrac{\pi}{4}\right)=\tan\!\left(x-\tfrac{\pi}{4}\right)+\sec^{2}\!\left(x-\tfrac{\pi}{4}\right)

Step 4: This is the

e^{x}(f+f')

form with

f(x)=\tan\!\left(x-\dfrac{\pi}{4}\right),\ f'(x)=\sec^{2}\!\left(x-\dfrac{\pi}{4}\right)

I=\int e^{x}\big(f+f'\big)\,dx=e^{x}\tan\!\left(x-\dfrac{\pi}{4}\right)+C

Correct answer: (2)

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