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

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

Answers (1)

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}

 

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