Question

int (cos4x/sin2x) dx​

Answer 1

[math]\cos (4x) = 1–2\sin^2 (2x) \text{, and: } \sin(2x) = 2\sin x \cos x[/math]

The given fraction F becomes:

[math]F=\dfrac{1}{\sin(2x)} - 2\sin(2x)=\dfrac{1}{2\sin x \cos x} -4 \sin x \cos x=[/math]

[math]=\dfrac{1}{2}(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}) -4 \sin x \cos x[/math]

Keeping in mind that the derivative of sin is cos, and the derivative of cos is (-sin), that integrates to:

[math]I= \dfrac{1}{2}(-ln|\cos x|+ ln|\sin x|)- 2\sin^2 x + C =[/math]

[math]= \dfrac{1}{2}ln|\tan x|- 2\sin^2 x + C[/math]

To test this result, we can differentiate the above, to get:

[math]I’=\dfrac{1}{2}[(\dfrac{1}{\tan x})(\sec^2 x)]- 4 \sin x \cos x=[/math]

[math]=\dfrac{1}{2}(\dfrac{\cos x}{\sin x}) \dfrac{1}{\cos ^2 x}- 2\sin(2x)=[/math]

[math]=\dfrac{1}{2\sin x \cos x}-2\sin (2x)=\dfrac{1}{\sin (2x)}-2\sin(2x)[/math]

…which is pretty much where we started from (fraction F).