#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 8 Maths Textbook Solution.

Answer: $-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cot 3 x+C$

Hint :-   Use substitution method to solve this integral.

Given: $\int \operatorname{cosec}^{4} 3 x d x$

Solution:Let $\mathrm{I}=\int \operatorname{cosec}^{4} 3 x d x$

Re-Write $\mathrm{I}=\int \operatorname{cosec}^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x d x$

$\left.I=\int\left(1+\cot ^{2} 3 x\right) \cdot \operatorname{cosec}^{2} 3 x d x \: \: \: \: \: \: \: \: \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$

\begin{aligned} &I=\int\left(\operatorname{cosec}^{2} 3 x+\cot ^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x\right) d x \\ &I=\int \operatorname{cosec}^{2} 3 x \cdot \cot ^{2} 3 x d x+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}

Substitute,  cot3x = t

\begin{aligned} &\operatorname{cosec}^{2} 3 x \cdot 3 d x=d t, \text { then } \\ &I=\int t^{2} \frac{d t}{3}+\int \operatorname{cosec}^{2} 3 x d x \\ &=-\frac{1}{3} \int t^{2} d t+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}

$=-\frac{1}{3} \frac{t^{2+1}}{2+1}+\frac{-\cot 3 x}{3}+C$                                    $\text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \operatorname{cosec}^{2} x d x=-\cot x+c$

$=-\frac{1}{3} \cdot \frac{t^{3}}{3}-\frac{\cot 3 x}{3}+C$

$=-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cdot \cot 3 x+\mathrm{C} \: \: \: \: \: \: \: \: \quad(\text { if, } \cot 3 \mathrm{x}=\mathrm{t})$