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#### Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 83 Maths Textbook Solution.

$-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)+c$

Hint:

You must know about integration of sec x & cosec x.

Given:

$\int \frac{1}{\sec x+\operatorname{cosec} x} d x$

Soltuion:

$\operatorname{now}, \int \frac{1}{\frac{1}{\cos x}+\frac{1}{\sin x}} d x \quad\left[\because \sec x=\frac{1}{\cos x}, \operatorname{cosec} x=\frac{1}{\sin x}\right]$

$=\int \frac{d x(\cos x \sin x)}{\sin x+\cos x}$

$=\frac{1}{2} \int \frac{2(\cos x \sin x)}{\sin x+\cos x} d x$                                $\left[\because(1+2 \sin x \cos x)=(\sin x+\cos x)^{2}\right.$

$=\frac{1}{2} \int \frac{1+2 \sin x \cos x-1}{\sin x+\cos x} d x$

$=\frac{1}{2} \int \frac{(\sin x+\cos x)^{2}-1}{\sin x+\cos x} d x$

$=\frac{1}{2} \int(\sin x+\cos x)=d x-\int \frac{1}{\sin x+\cos x} d x$

$=\frac{1}{2}\left[\int \sin x d x+\int \cos x d x-\int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)}\right]$

$=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(x+\left(\frac{\pi}{4}\right)\right.}\right]$

$=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \int \operatorname{cosec}\left(x+\frac{\pi}{4}\right) d x\right]$

$=-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)$

$=-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)+c$