#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 11 Maths Textbook Solution.

Answer: $-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C$

Hint :- Use substitution method to solve this integral.

Given: $\int \cot ^{5} x d x$

Solution: Let $I=\int \cot ^{5} x d x$

Re-Write,$\mathrm{I}=\int \cot ^{5} x d x$

$=\int \cot ^{3} x \cdot \cot ^{2} x d x$

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

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

\begin{aligned} &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x-1\right) \cot x d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x \cdot \cot x-\cot x\right) d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \operatorname{cosec}^{2} x \cdot \cot x d x+\int \cot x d x \end{aligned}

Substitute,  cotx = t

\begin{aligned} &-\operatorname{cosec}^{2} x d x=d t, \text { then } \\ &I=\int t^{3}(-d t)-\int t(-d t)+\int \cot x d x \\ &=-\int t^{3} d t+\int t d t+\int \cot x d x \end{aligned}

$=-\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sin x|+C$                                                            $\inline \text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c} \& \int \cot x d x=\log |\sin x|+c\right)$

\inline \begin{aligned} &=-\frac{1}{4} t^{4}+\frac{1}{2} t^{2}+\log |\sin x|+C \\ &=-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C \end{aligned}                                                        $\inline \left ( as,t=\cot x \right )$