Question
Evaluate the following:
$\int\frac{\sin^6\text{x}+\cos^6\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$
$\int\frac{\sin^6\text{x}+\cos^6\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$
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Answer
Let $\text{I}=\int\frac{\sin^6\text{x}+\cos^6\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$ $=\int\frac{(\sin^2\text{x})^3+(\cos^2\text{x})^3}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\frac{(\sin^2\text{x}+\cos^2\text{x})(\sin^4\text{x}-\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x})}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\frac{\sin^4\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}+\int\frac{\cos^4\text{x}}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}-\int\frac{\sin^2\text{x}\cos^2\text{x}}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{xdx}+\int\cos^2\text{xdx}-\int1\text{dx}$
$=\int(\sec^2\text{x}-1)\text{dx}+\int(\cos\text{ec}^2\text{x}-1)\text{dx}-\int1\text{dx}$
$=\int\sec^2\text{xdx}+\int\cos\text{ec}^2\text{xdx}-3\int\text{dx}$
$\text{I}=\tan\text{x}-\cot\text{x}-3\text{x}+\text{C}$
$=\int\frac{(\sin^2\text{x}+\cos^2\text{x})(\sin^4\text{x}-\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x})}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\frac{\sin^4\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}+\int\frac{\cos^4\text{x}}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}-\int\frac{\sin^2\text{x}\cos^2\text{x}}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{xdx}+\int\cos^2\text{xdx}-\int1\text{dx}$
$=\int(\sec^2\text{x}-1)\text{dx}+\int(\cos\text{ec}^2\text{x}-1)\text{dx}-\int1\text{dx}$
$=\int\sec^2\text{xdx}+\int\cos\text{ec}^2\text{xdx}-3\int\text{dx}$
$\text{I}=\tan\text{x}-\cot\text{x}-3\text{x}+\text{C}$
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