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

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

Answers (1)

Answer:
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

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