provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 26

Answer: $\frac{-1}{\sin x+\cos x}+c$

Hint:Use substitution method to solve this integral.

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

Solution:

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

$=\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+\sin 2 x} d x \quad\left[\because 1=\sin ^{2} x+\cos ^{2} x\right]$

$=\int \frac{\cos x-\sin x}{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} d x \quad[\because \sin 2 x=2 \sin x \cos x]$

$=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x \quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$

$\text { Put } \sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t \text { then }$

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

$=\int \frac{1}{t^{2}}=\int t^{-2} d t=\frac{t^{-2+1}}{-2+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

$=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c$

$=\frac{-1}{\sin x+\cos x}+c \quad[\because t=\sin x+\cos x]$