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Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 2 maths textbook solution.

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Answer :  0

Given : \int_{0}^{2 \pi} \log (\sec x+\tan x) d x

Hint : You must know about the trignometric identities

Solution : I=\int_{0}^{2 \pi} \log (\sec x+\tan x) d x-------(1)

We know that

\begin{aligned} &\int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x \\ &I=\int_{0}^{2 \pi} \log (\sec (2 \pi-x)+\tan (2 \pi-x)) d x \\ &I=\int_{0}^{2 \pi} \log (\sec x-\tan x) d x \quad-------(2)[\sec (2 \pi-x)=\sec x, \tan (2 \pi-x)=-\tan x] \end{aligned}

Add (1) and (2)

\begin{aligned} &2 I=\int_{0}^{2 \pi} \log (\sec x+\tan x)+\log (\sec x-\tan x) d x \\ &2 I=\int_{0}^{2 \pi} \log (1) d x\left[\tan ^{2} x+1=\sec ^{2} x\right] \\ &2 I=0[\log (1)=0] \\ &I=0 \end{aligned}

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