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How to integrate :

$$\int \frac{\sin^4x+\cos^4x}{\sin x \cos x}\:dx$$

$$=\int \:\sin^2x \tan x \: dx+\int \:\cos^2x \cot x \:dx$$

Any suggestion?

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  • $\begingroup$ If you're going to submit an edit, please make sure that it is accurate and your MathJax works. This way, we can avoid the edit war that happened on this post. $\endgroup$ – Michael Morrow Jun 19 '20 at 3:48
  • $\begingroup$ I have to be nitpicky and point out that you integrate a function or evaluate an integral (unless you’re computing a double integral). $\endgroup$ – gen-ℤ ready to perish Jun 19 '20 at 4:49
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Add and subtract in nominator $2 \sin^2x \cos^2x$. Can you continue?

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  • $\begingroup$ Yes,I can.@zkutch Thank you! $\endgroup$ – ikigai Jun 19 '20 at 3:49
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$$\int \frac{\sin^4x+\cos^4x}{\sin x \cos x}\:dx$$ $$=\int \frac{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{\sin x \cos x}\:dx$$ $$=\int \frac{1-2\sin^2x\cos^2x}{\sin x \cos x}\:dx$$ $$=\int (2\csc 2x-\sin 2x)\:dx$$

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An alternative approach is to write the integral as $\int\frac{\sin^3xdx}{\cos x}+\int\frac{\cos^3xdx}{\sin x}$. In the first part, use $u=\cos x$ to get $\int\frac{(u^2-1)du}{u}=\frac12u^2-\ln|u|+C$, where $C$ is a locally constant function that can change whenever $u=0$, i.e. at $x\in\pi\Bbb Z\setminus\tfrac{\pi}{2}\Bbb Z$. In the second part, use $v=\sin x$ to get $\int\frac{(1-v^2)dv}{v}=\ln|v|-\frac12v^2+C^\prime$, with $C^\prime$ locally constant but able to change at $x\in\pi\Bbb Z$. So$$\int\frac{\sin^4x+\cos^4x}{\sin x\cos x}dx=\ln|\tan x|+\frac12(\cos^2x-\sin^2x)+K,$$where $K$ is locally constant but can change at $x\in\tfrac{\pi}{2}\Bbb Z$.

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