# How to integrate :$\int \frac{\sin^4x+\cos^4x}{\sin x \cos x}\:dx$

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?

• 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. – Michael Morrow Jun 19 '20 at 3:48
• I have to be nitpicky and point out that you integrate a function or evaluate an integral (unless you’re computing a double integral). – gen-ℤ ready to perish Jun 19 '20 at 4:49

Add and subtract in nominator $$2 \sin^2x \cos^2x$$. Can you continue?
$$\int \frac{\sin^4x+\cos^4x}{\sin x \cos x}\:dx$$ $$=\int \frac{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{\sin x \cos x}\:dx$$ $$=\int \frac{1-2\sin^2x\cos^2x}{\sin x \cos x}\:dx$$ $$=\int (2\csc 2x-\sin 2x)\:dx$$
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 $$x\in\tfrac{\pi}{2}\Bbb Z$$.