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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 7

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Answer:
The correct answer is e^{x} \log \sec x+c
 

 

Given:

\int e^{x}(\tan x-\log \cos x) d x

Solution:

    I=\int e^{x}(\tan x-\log \cos x) d x

        =\int e^{x} \tan x\; d x-\int e^{x} \log \cos x \; d x

Integrating by parts,

        \begin{aligned} &=\int e^{x} \tan x \; d x-e^{x} \log \cos x+\int e^{x}\left(\frac{d}{d x} \log \cos x\right) d x \\ &=\int e^{x} \tan x \; d x-e^{x} \log \cos x-\int e^{x} \tan x \; d x \end{aligned}

        \begin{aligned} &=-e^{x} \log \cos x+\mathrm{c} \\ &=e^{x} \log \sec x+c \end{aligned}

So, the correct answer is e^{x} \log \sec x+c.

 

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