explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 32 maths

Answer:$\frac{1}{2}\{\log (\operatorname{cosec} x-\cot x)\}^{2}+c$

Hint:Use substitution method to solve this integral.

Given:   $\int \operatorname{cosec} x \cdot \log (\operatorname{cosec} x-\cot x) d x$

Solution:

\begin{aligned} &\text { Let } I=\int \operatorname{cosec} x \cdot \log (\operatorname{cosec} x-\cot x) d x \\ &\text { Put } \log (\operatorname{cosec} x-\cot x)=t \end{aligned}

$\Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)}\left(-\operatorname{cosec} x \cdot \cot x-\left(-\operatorname{cosec}^{2} x\right)\right) d x=d t$

$\Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)}\left\{\operatorname{cosec}^{2} x-\operatorname{cosec} x \cdot \cot x\right\} d x=d t$

$\Rightarrow \frac{1}{(\operatorname{cosec} x-\cot x)} \operatorname{cosec} x\{\operatorname{cosec} x-\cot x\} d x=d t$

$\Rightarrow \operatorname{cosec} x d x=d t \Rightarrow d x=\frac{d t}{\operatorname{cosec} x} \text { then }$

$\Rightarrow I=\int \operatorname{cosec} x \cdot t \cdot \frac{d t}{\operatorname{cosec} x}=\int t \; d t$

$=\frac{t^{1+1}}{1+1}+c=\frac{t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c,\right]$

$=\frac{1}{2}\{\log (\operatorname{cosec} x-\cot x)\}^{2}+c \quad \quad[\because t=\log (\operatorname{cosec} x-\cot x)]$