#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 10 Maths Textbook Solution.

Answer: $-\frac{1}{6} \cot ^{6} x-\frac{1}{8} \cot ^{8} x+C$

Hint :-  Use substitution method to solve this integral.

Given:$\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$

Solution:Let, $\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$

Re-Write,$\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{2} x \cdot \operatorname{cosec}^{2} x d x$

$\mathrm{I}=\int \cot ^{5} x\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x$            $\text { (if, } \left.\operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$

$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{5} x \cdot \cot ^{2}\right) \operatorname{cosec}^{2} x d x$

$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{7} x\right) \operatorname{cosec}^{2} x d x$

Substitute,  cotx = t

$-\operatorname{cosec}^{2} x d x=d t, \text { then }$

$I=\int\left(t^{5}+t^{7}\right) \operatorname{cosec}^{2} x \cdot \frac{d t}{-\operatorname{cosec}^{2} x} \mathrm{dx} \: \: \: \: \: \: \: \: \quad(\text { as } \cot x=t)$

\begin{aligned} &=-\int\left(t^{5}+t^{7}\right) \mathrm{dt} \\ &=-\int t^{5} d t-\int t^{7} \mathrm{dt} \end{aligned}

$=-\frac{t^{5+1}}{5+1}-\frac{t^{7+1}}{7+1}+C$                $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c}\right)$

$=-\frac{t^{6}}{6}-\frac{t^{8}}{8}+C$

$=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+\mathrm{C}$                    $(\text { if, } t=\cot x)$