Definite IntegrationmediumFree

∫(πx−4x²)ln(1+tanx) dx, 0 to π/4 | JEE

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

Question

\displaystyle\int_{0}^{\pi/4}\big(\pi x-4x^{2}\big)\ln(1+\tan x)\,dx

equals

\dfrac{\pi^{3}}{192}

\dfrac{\pi^{3}}{192}\ln 2

correct

\dfrac{\pi^{2}}{96}

\dfrac{\pi^{2}}{96}\ln 2

Solution

Step 1: Write the polynomial factor in a symmetric form. Since

\pi x-4x^{2}=4x\left(\dfrac{\pi}{4}-x\right)

I=\int_{0}^{\pi/4}4x\left(\dfrac{\pi}{4}-x\right)\ln(1+\tan x)\,dx.

Step 2: Apply the King's property

\displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx

with

a=\dfrac{\pi}{4}

. The factor

4x\left(\dfrac{\pi}{4}-x\right)

is symmetric under

x\mapsto\dfrac{\pi}{4}-x

, while for the logarithm we use the classic identity

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

Hence

I=\int_{0}^{\pi/4}4x\left(\dfrac{\pi}{4}-x\right)\big[\ln 2-\ln(1+\tan x)\big]\,dx=4\ln 2\int_{0}^{\pi/4}x\left(\dfrac{\pi}{4}-x\right)dx-I.

Step 3: Solve for

I

2I=4\ln 2\int_{0}^{\pi/4}x\left(\dfrac{\pi}{4}-x\right)dx\ \Rightarrow\ I=2\ln 2\int_{0}^{\pi/4}\left(\dfrac{\pi}{4}x-x^{2}\right)dx.

Step 4: Evaluate the elementary integral:

\int_{0}^{\pi/4}\left(\dfrac{\pi}{4}x-x^{2}\right)dx=\left[\dfrac{\pi}{8}x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{\pi/4}=\dfrac{\pi}{8}\cdot\dfrac{\pi^{2}}{16}-\dfrac{1}{3}\cdot\dfrac{\pi^{3}}{64}=\dfrac{\pi^{3}}{128}-\dfrac{\pi^{3}}{192}=\dfrac{\pi^{3}}{384}.

Therefore

I=2\ln 2\cdot\dfrac{\pi^{3}}{384}=\dfrac{\pi^{3}}{192}\ln 2.

Correct answer: (2)

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