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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 31

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Answer: \frac{1}{2}\{\log (\sec x+\tan x)\}^{2}+c

Hint:Use substitution method to solve this integral.

Given:   \int \sec x \cdot \log (\sec x+\tan x) d x


        \begin{aligned} &\text { Let } I=\int \sec x \cdot \log (\sec x+\tan x) d x \\ &\text { Put } \log (\sec x+\tan x)=t \end{aligned}

        \begin{aligned} &\Rightarrow \frac{1}{(\sec x+\tan x)}\left(\sec x \tan x+\sec ^{2} x\right) d x=d t \\ &\Rightarrow \frac{1}{(\sec x+\tan x)} \sec x(\tan x+\sec x) d x=d t \end{aligned}

        \begin{aligned} &\Rightarrow \sec x\; d x=d t \Rightarrow d x=\frac{d t}{\sec x} \text { then } \\ &\Rightarrow I=\int \sec x \cdot t \cdot \frac{d t}{\sec x}=\int t \; d t \end{aligned}

        =\frac{t^{1+1}}{1+1}+c=\frac{t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c,\right]

        =\frac{1}{2}\{\log (\sec x+\tan x)\}^{2}+c \quad[\because t=\log (\sec x+\tan x)]

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