# Types of trigonometric integrals

I am trying to solve the following integral: $$\int \dfrac{\cos^3(x)}{\sin^2(x)+\sin^4(x)}\,dx.$$

After many attempts (used even Wolfram-alpha it suggested I should multiply by $\sec^4(x)$ and pull a magic $u$ substitution...) I wrote $\cos^3(x) = \cos x (1-\sin^2(x))$ and from there substituted the $\sin$ and did partial fractions etc.

But I wonder what if the power of $\cos$ is even? For example let's say you have the integral: $$\int \dfrac{\cos^4(x)}{\sin^2(x)+\sin^4(x)}dx$$

Hint:

Divide numerator and denominator by $\cos^4x$

$$I=\int\dfrac{\cos^4x}{\sin^2x+\sin^4x}dx=\int\dfrac1{\tan^2x(\tan^2x+\sec^2x)}dx$$

Set $\tan x=u$

$$I=\int\dfrac{du}{u^2(2u^2+1)(u^2+1)}$$

Set $u^2=v$

$$\dfrac1{v(2v+1)(v+1)}=\dfrac Av+\dfrac B{2v+1}+\dfrac C{v+1}$$

Replace $\cos^2x$ by $(1+\cos2x)/2$, and $\sin^2x$ by $(1-\cos2x)/2$.

• Without computing it, I'll get a function of $cos(2x)$ but with no derivative of it... So how do i move on? Feb 5 '18 at 17:45
• Combine that with Giuseppe Negro's tan(x/2) hint, and you get lab's u=tan x substitution. Feb 5 '18 at 18:32