Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 13 Maths Textbook Solution.

Answer:  $log\left ( \frac{\sqrt{2}-1}{2-\sqrt{3}} \right )$

Hint: Use indefinite formula then put the limit to get the required answer

Given:$\int_{\frac{\pi }{6}}^{\frac{\pi }{4}}\cos ecxdx$

Solution:$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x=[\log |\cos e c x-\cot x|]_{\frac{\pi}{6}}^{\frac{\pi}{4}}\: \: \: \: \: \: \: \: \: \quad\left[\int \cos \operatorname{ecx} d x=\log |\operatorname{cosec} x-\cot x|\right]$

$=\left[\log \left|\cos e c \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|\operatorname{cosec} \frac{\pi}{6}-\cot \frac{\pi}{6}\right|\right]\left[\begin{array}{l} \cos e c \frac{\pi}{4}=\sqrt{2} \\ \cot \frac{\pi}{4}=2 \\ \cos \operatorname{ec} \frac{\pi}{6}=2 \\ \cot \frac{\pi}{6}=\sqrt{3} \end{array}\right]$

\begin{aligned} &=[\log (\sqrt{2}-1)-\log (2-\sqrt{3})] \\ &=\log (\sqrt{2}-1)-\log (2-\sqrt{3}) \end{aligned} \quad\left[\log a-\log b=\log \frac{a}{b}\right]

$=log\left ( \frac{\sqrt{2}-1}{2-\sqrt{3}} \right )$