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Mathematics Question on Some Properties of Definite Integrals

The value of the integral \int_\limits{-\pi/4}^{\pi/4} \log (\sec \theta-\tan \theta) d \theta is

A

π4\frac {\pi}{4}

B

π2\frac {\pi}{2}

C

00

D

π\pi

Answer

00

Explanation

Solution

Let I=\int_\limits{-\pi / 4}^{\pi / 4} \log (\sec \theta-\tan \theta) d \theta
Again, let f(θ)=log⁡(sec⁡θ−tan⁡θ)f(\theta)=\log (\sec \theta-\tan \theta)
∴f(−θ)=log⁡[sec⁡(−θ)−tan⁡(−θ)]\therefore f(-\theta)=\log [\sec (-\theta)-\tan (-\theta)]
=log⁡[(sec⁡θ+tan⁡θ)×sec⁡θ−tan⁡θsec⁡θ−tan⁡θ]=\log \left[(\sec \theta+\tan \theta) \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}\right]
=log⁡[sec⁡2θ−tan⁡2θsec⁡θ−tan⁡θ]=log⁡[1sec⁡θ−tan⁡θ]=\log \left[\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta-\tan \theta}\right]=\log \left[\frac{1}{\sec \theta-\tan \theta}\right]
=log⁡1−log⁡(sec⁡θ−tan⁡θ)=\log 1-\log (\sec \theta-\tan \theta)
=0−log⁡(sec⁡θ−tan⁡θ)=0-\log (\sec \theta-\tan \theta)
⇒f(−θ)=−f(θ)\Rightarrow f(-\theta)=-f(\theta)
Hence, f(θ)f(\theta) is an odd function.
∴I=0\therefore I=0