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

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

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

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

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

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