1
$\begingroup$

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

$\endgroup$
1
1
$\begingroup$

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 $u^2=v$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.