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Official PYQ · 2009Q25Maxima and Minima, Mean Value Theorem, IntegrationMCQ1 mark

GATE CS 2009 Q25 — Maxima and Minima, Mean Value Theorem, Integration

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∫0π/4(1−tan⁡x)/(1+tan⁡x) dx\int^{\pi/4}_0 (1-\tan x)/(1+\tan x)\,dx

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Answer: D

We want to evaluate the definite integral:

I=∫0π/41−tan⁡x1+tan⁡x dxI = \int_{0}^{\pi/4} \frac{1-\tan x}{1+\tan x} \, dx

Using the trigonometric identity tan⁡(π4−x)=tan⁡(π/4)−tan⁡x1+tan⁡(π/4)tan⁡x=1−tan⁡x1+tan⁡x\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4)\tan x} = \frac{1-\tan x}{1+\tan x}, we can rewrite the integrand:

I=∫0π/4tan⁡(π4−x) dxI = \int_{0}^{\pi/4} \tan\left(\frac{\pi}{4} - x\right) \, dx

Let u=π4−xu = \frac{\pi}{4} - x, which gives du=−dxdu = -dx. Changing the limits of integration:

  • When x=0x = 0, u=π4u = \frac{\pi}{4}
  • When x=π4x = \frac{\pi}{4}, u=0u = 0

Substituting these into the integral:

I=∫π/40tan⁡(u)(−du)=∫0π/4tan⁡(u) duI = \int_{\pi/4}^{0} \tan(u) (-du) = \int_{0}^{\pi/4} \tan(u) \, du

Since ∫tan⁡(u) du=ln⁡∣sec⁡u∣\int \tan(u) \, du = \ln|\sec u|, we evaluate the limits:

I=[ln⁡∣sec⁡u∣]0π/4=ln⁡(sec⁡π4)−ln⁡(sec⁡0)I = \Big[ \ln|\sec u| \Big]_{0}^{\pi/4} = \ln\left(\sec\frac{\pi}{4}\right) - \ln(\sec 0) I=ln⁡(2)−ln⁡(1)=ln⁡(21/2)−0=12ln⁡2I = \ln(\sqrt{2}) - \ln(1) = \ln\left(2^{1/2}\right) - 0 = \frac{1}{2}\ln 2

Thus, the correct option is D.

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