#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 16 Maths Textbook Solution.

Answer: $\frac{-1}{8} \cos 4 x+c$

Hint: $\text { To solve this equation we use } \int \sin 2 x_{5} \int \cos 2 x d x \text { formula }$

Given: $\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$

Solution: $\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$

\begin{aligned} &=\frac{2 \cos ^{2} 2 x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}} d x \\ &=\frac{2 \cos ^{2} 2 x}{\frac{\cos ^{2} x-\sin ^{2} x}{\sin x \cos x}} d x \\ &=\frac{2(\sin x \cos x)(\cos 2 x)^{2}}{\cos ^{2} x-\sin ^{2} x} d x \end{aligned}

\begin{aligned} &=\frac{\sin 2 x(\cos 2 x)^{2}}{\cos 2 x} \\ &=\sin 2 x \cos 2 x \end{aligned} \quad\left[\begin{array}{c} \because 2 \sin x \cos x=\sin 2 x \\ \cos ^{2} x-\sin ^{2} x=\cos 2 x \end{array}\right]

\begin{aligned} &\text { Multiply } 2 \text { in numerator and denominator }\\ &=\frac{1}{2}(2 \sin 2 x \cos 2 x)\\ &=\frac{1}{2}(\sin 4 x) \quad[\because 2 \sin 2 x \cos 2 x=\sin 4 x] \end{aligned}

\begin{aligned} &\int \frac{1+\cos 4 x}{\cot x-\tan x} d x=\int \frac{1}{2}(\sin 4 x) d x \\ &=\frac{1}{2} \int(\sin 4 x) d x \quad\left[\because \int \sin (a x+b) d x=\frac{-1}{a} \cos (a x+b)+c, a \neq 0\right] \\ &=\frac{1}{2}\left(\frac{-1}{4} \cos 4 x\right)+c \\ &=\frac{-1}{8} \cos 4 x+c \end{aligned}