# Get Answers to all your Questions

### 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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