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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 12 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 12 Maths Textbook Solution.

Answers (1)

Answer:log\left ( \sqrt{2}+1 \right )

Hint:: Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given: \int_{0}^{\frac{\pi }{4}}\sec xdx

Solution: \int_{0}^{\frac{\pi}{4}} \sec x d x=[\log |\sec x+\tan x|]_{0}^{\frac{\pi}{4}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sec x d x=\log |\sec x+\tan x|\right]

\left.\begin{array}{l} =\left[\log \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|-\log |\sec 0-\tan 0|\right] \\ =[\log |\sqrt{2}+1|-\log |1-0|] \end{array}\right]\left[\begin{array}{l} \sec \frac{\pi}{4}=\sqrt{2} \\ \tan \frac{\pi}{4}=1 \ \\ \sec 0=1 \\ \tan 0=0 \end{array}\right]

=log\left ( \sqrt{2}+1 \right )-log1

=log\left ( \sqrt{2}+1 \right )-0                                                                                                        \left [ log1=0 \right ]

=log\left ( \sqrt{2}+1 \right )

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