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Question

$\int\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\text{dx}$

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Answer

$\int\Big(\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{(1+\cos4\text{x})}{\big(\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\big)}\text{dx}$
$=\int\frac{2\cos^22\text{x}\times\sin\text{x}\cos\text{x}}{(\cos^2\text{x}-\sin^2\text{x})}\text{dx}$
$=\int\frac{\cos^22\text{x}\times2\sin\text{x}\cos\text{x}}{\cos2\text{x}}\text{dx}$
$=\int\cos2\text{x}\sin2\text{x}\text{ dx}$
$=\frac{1}{2}\int2\sin2\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\int\sin4\text{x dx}$
$=\frac{1}{2}\Big[-\frac{\cos4\text{x}}{4}\Big]+\text{c}$
$=-\frac{1}{8}\cos4\text{x}+\text{c}$

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