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: $$\int \dfrac{\cos^4(x)}{\sin^2(x)+\sin^4(x)}dx$$



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


Set $\tan x=u$


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 $(1-\cos2x)/2$.

  • $\begingroup$ Without computing it, I'll get a function of $cos(2x)$ but with no derivative of it... So how do i move on? $\endgroup$
    – Erald Sn
    Feb 5 '18 at 17:45
  • $\begingroup$ Combine that with Giuseppe Negro's tan(x/2) hint, and you get lab's u=tan x substitution. $\endgroup$
    – Empy2
    Feb 5 '18 at 18:32

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