Indefinite IntegrationhardFree

Evaluate ∫dx/((tanx+1)sin²x) at π/4 | JEE

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

Question
∫dx(tan⁡x+1)sin⁡2x=f(x)+c\displaystyle\int \dfrac{dx}{(\tan x+1)\sin^{2}x}=f(x)+c, where f ⁣(π2)=0f\!\left(\dfrac{\pi}{2}\right)=0. Then f ⁣(π4)f\!\left(\dfrac{\pi}{4}\right) is
A1−ln⁡21-\ln 2
Bln⁡2−1\ln 2-1correct
C1+ln⁡21+\ln 2
D−1−ln⁡2-1-\ln 2
Solution
Step 1: Express 1sin⁡2x\dfrac{1}{\sin^{2}x} in terms of tan⁡x\tan x. Since 1sin⁡2x=sec⁡2xtan⁡2x\dfrac{1}{\sin^{2}x}=\dfrac{\sec^{2}x}{\tan^{2}x},
I=∫sec⁡2x(tan⁡x+1)tan⁡2x dxI=\int\dfrac{\sec^{2}x}{(\tan x+1)\tan^{2}x}\,dx
 Step 2: Substitute t=tan⁡x⇒dt=sec⁡2x dxt=\tan x\Rightarrow dt=\sec^{2}x\,dx:
I=∫dtt2(t+1)I=\int\dfrac{dt}{t^{2}(t+1)}
 Step 3: Partial fractions, 1t2(t+1)=At+Bt2+Dt+1\dfrac{1}{t^{2}(t+1)}=\dfrac{A}{t}+\dfrac{B}{t^{2}}+\dfrac{D}{t+1}, gives A=−1, B=1, D=1A=-1,\ B=1,\ D=1:
1t2(t+1)=1t2−1t+1t+1\dfrac{1}{t^{2}(t+1)}=\dfrac{1}{t^{2}}-\dfrac{1}{t}+\dfrac{1}{t+1}
 Step 4: Integrate and back-substitute t=tan⁡xt=\tan x (so 1t=cot⁡x\dfrac{1}{t}=\cot x and t+1t=1+cot⁡x\dfrac{t+1}{t}=1+\cot x):
I=−1t+ln⁡∣t+1t∣+c=−cot⁡x+ln⁡∣1+cot⁡x∣+cI=-\dfrac{1}{t}+\ln\left|\dfrac{t+1}{t}\right|+c=-\cot x+\ln|1+\cot x|+c
⇒f(x)=−cot⁡x+ln⁡∣1+cot⁡x∣+c\Rightarrow f(x)=-\cot x+\ln|1+\cot x|+c
 Step 5: Apply f ⁣(π2)=0f\!\left(\dfrac{\pi}{2}\right)=0. At x=π2x=\dfrac{\pi}{2}, cot⁡x=0\cot x=0, so f ⁣(π2)=0+ln⁡1+c=c=0f\!\left(\dfrac{\pi}{2}\right)=0+\ln 1+c=c=0. Step 6: Evaluate at x=π4x=\dfrac{\pi}{4}, where cot⁡x=1\cot x=1:
f ⁣(π4)=−1+ln⁡2=ln⁡2−1f\!\left(\dfrac{\pi}{4}\right)=-1+\ln 2=\ln 2-1
 Correct answer: (2)
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