Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 13 Marks
Question
Evaluate $\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x$
✓
Answer
According to the question,
$I=\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x$
$=\int e^{2 x} (\frac{1-2 \sin x \cos x}{2 \sin ^2 x}) d x\ [\because 1-\cos 2 x=2 \sin ^2 x$ and $\sin 2 x=2 \sin x \cos x]$
$=\frac{1}{2} \int e^{2 x}\left(\operatorname{cosec}^2 x-2 \cot x\right) d x$
$=\frac{1}{2} \int e_I^{2 x} \operatorname{cosec}_{I I}^2 x d x-\int e^{2 x} \cot x d x $
Using integration by parts for first integral :
$ =\frac{1}{2}\left[e^{2 x} \int \operatorname{cosec} e^2 x d x-\int\left\{\frac{d}{d x}\left(e^{2 x}\right) \int \operatorname{cosec} e^2 x d x\right\} d x\right]-\int e^{2 x} \cot x d x$
$=\frac{1}{2}\left[-e^{2 x} \cot x+\int 2 e^{2 x} \cot x d x\right]+C-\int e^{2 x} \cot x d x$
$=-\frac{e^{2 x}}{2} \cot x+\int e^{2 x} \cot x d x-\int e^{2 x} \cot x d x+C$
$I=-\frac{e^{2 z}}{2} \cot x+C$
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