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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate: $\int\text{e}^{x}\Bigg(\frac{\sin 4x - 4}{1-\cos 4x}\Bigg)\text{dx}$.
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Answer

$\text{I}=\int\text{e}^{x}\Bigg(\frac{\sin 4x - 4}{1-\cos 4x}\Bigg)\text{dx}$
$=\int\text{e}^{x}\Bigg[\frac{\sin 4x}{1-\cos 4x}-\frac{4}{1 - \cos 4x}\Bigg]\text{dx}$
$=\int\text{e}^{x}\Bigg[\frac{2\sin 2x \cos 2x}{2\cdot\sin^{2} 2x}-\frac{4}{2\sin^{2}2x}\Bigg]\text{dx}$
$=\int\text{e}^{x}[\cot2x-2\text{cosec}^{2}2x]\text{dx}$
This is of the form $=\int\text{e}^{x}[\text{f(x)+f'(x)}]\text{dx}$
$\therefore \text{I}=\text{e}^{x}\cot\text{2x + c}.$

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