Question
Evalute the following integrals:
$\int\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\text{dx}$
$\int\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\text{dx}$
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Answer
$\int\Big(\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\Big)\text{dx}$
$=\int\frac{-2\sin\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}{2\cos\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}\text{dx}$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\ \& \\ \sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\bigg]$
$=-\int\frac{\sin3\text{x}}{\cos3\text{x}}\text{dx}$
$=-\int\tan3\text{x dx}$
$=\frac{-\text{In}|\sec3\text{x}|}{3}+\text{C}$
$=\frac{1}{3}\text{ln}\big(|\sec3\text{x}|\big)^{-1}+\text{C}$
$=\frac{1}{3}\text{ln}|\cos3\text{x}|+\text{C}$
$=\int\frac{-2\sin\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}{2\cos\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}\text{dx}$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\ \& \\ \sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\bigg]$
$=-\int\frac{\sin3\text{x}}{\cos3\text{x}}\text{dx}$
$=-\int\tan3\text{x dx}$
$=\frac{-\text{In}|\sec3\text{x}|}{3}+\text{C}$
$=\frac{1}{3}\text{ln}\big(|\sec3\text{x}|\big)^{-1}+\text{C}$
$=\frac{1}{3}\text{ln}|\cos3\text{x}|+\text{C}$
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