#### Provide Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 8 Maths Textbook Solution.

Answer : $\mathrm{I}=\frac{-1}{\sqrt{2}} \log \left|\cos e c\left(x-\frac{\pi}{4}\right)+\cot \left(x-\frac{\pi}{4}\right)\right|+c$

Hint: To solve this equation we have to use the formula of $\cos A-\sin B$

Given:  $\int \frac{1}{\cos x-\sin x} d x$

Solution :

\begin{aligned} &\int \frac{(\sqrt{2})}{\sqrt{2}(\cos x-\sin x)} d x \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x} d x \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sin \frac{\pi}{4} \cos x-\sin x \cos \frac{\pi}{4}} d x \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{2}} \int \frac{1}{\sin \left(\frac{\pi}{4}-x\right)} d x \; \; \; \; \; \; \; \; \; \; \; \quad[\sin a \cos b-\cos a \sin b=\sin (a-b)] \\ &=\frac{-1}{\sqrt{2}} \int \frac{1}{\sin \left(x-\frac{\pi}{4}\right)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; [\sin (-x)=-\sin x] \end{aligned}

\begin{aligned} &=\frac{-1}{\sqrt{2}} \int \cos e c\left(x-\frac{\pi}{4}\right) d x \\ &=\frac{-1}{\sqrt{2}} \log \left|\cos e c\left(x-\frac{\pi}{4}\right)+\cot \left(x-\frac{\pi}{4}\right)\right|+c \end{aligned}

Note : Final answer is not matching.