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  • need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 11

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 11

Answers (1)

Answer:
The correct answer is e^{x} \cot (2 x)+c
Hint:

\begin{aligned} i.\; \; \; \; \; &\cos 2 a=1-\sin ^{2} A \\ ii.\; \; \; \; \; &\sin 2 a=2 \sin A \cos A \end{aligned}

Given:  \int e^{x}\left(\frac{\sin 4 x-4}{1-\cos 4 x}\right) d x

Solution:

        I=\int e^{x}\left(\frac{\sin 4 x-4}{1-\cos 4 x}\right) d x

        \mathrm{I}=\int e^{x}\left(\frac{2 \sin (2 x) \cos (2 x)-4}{2 \sin ^{2}(2 x)}\right) d x \quad \therefore\left[\cos 2 a=1-\sin ^{2} A, \sin 2 a=2 \sin A \cos A\right]

        \mathrm{I}=\int e^{x}\left(\frac{2 \sin (2 x) \cos (2 x)}{2 \sin ^{2}(2 x)}-\frac{4}{2 \sin ^{2}(2 x)}\right) d x

            \Rightarrow \int e^{x}\left(\cot (2 x)-2 \operatorname{cosec}^{2}(2 x)\right) d x

\text { Where } f(x)=\cot (2 x) \text { and } f^{\prime}(x)=-\operatorname{2cosec}^{2}(2 x)

        \begin{aligned} &I=\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x \\ &=e^{x} f(x)+c \\ &=e^{x} \cot (2 x)+c \end{aligned}

So, the correct answer is e^{x} \cot (2 x)+c

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