#### explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 24 maths

The correct answer is  $-\frac{1}{2} e^{2 x} \cot x+c$
Hint:

$\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$

Given:

$\int e^{2 x} \frac{(1-\sin 2 x)}{(1-\cos 2 x)} d x$

Solution:

$I=\int e^{2 x} \frac{(1-\sin 2 x)}{(1-\cos 2 x)} d x$

\begin{aligned} &\text { Let } 2 x=t \\ &\operatorname{Or} 2 d x=d t \end{aligned}

\begin{aligned} &I=\int e^{t} \frac{(1-\sin t)}{(1-\cos t)} \frac{d t}{2} \\ &=\frac{1}{2} \int e^{t} \frac{\left(1-2 \sin \frac{t}{2} \cos \frac{t}{2}\right)}{2 \sin ^{2} \frac{t}{2}} d t \end{aligned}

\begin{aligned} &=\frac{1}{4} \int e^{t}\left(\operatorname{cosec}^{2} \frac{t}{2}-2 \cot \frac{t}{2}\right) d t \\ &=\frac{1}{4} \int e^{t}\left(-2 \cot \frac{t}{2}+\operatorname{cosec}^{2} \frac{t}{2}\right) d t \end{aligned}

\begin{aligned} &\text { Let } f(x)=-2 \cot \frac{t}{2} \\ &\text { Or } f^{\prime}(x)=\operatorname{cosec}^{2} \frac{t}{2} d t \end{aligned}

$\left[\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\right]$

\begin{aligned} &\therefore \quad I=\frac{1}{4} e^{t}\left(-2 \cot \frac{t}{2}+c\right) \\ &\operatorname{Or} I=-\frac{1}{2} e^{2 x} \cot x+c \end{aligned}

So, the correct answer is $-\frac{1}{2} e^{2 x} \cot x+c$