2
$\begingroup$

I want to find the integral $$\int{\dfrac{\cos^2 x}{\sin^2 x + 4\sin x \cos x}} \mathrm{d} x.$$ I tried put $t=\tan \frac{x}{2}$, I got $$\int{-\dfrac{1}{4}\dfrac{(t^2-1)^2}{t(t^2 + 1)(2t^2-t-2)} \mathrm{d} t.}$$ But, it's too difficult for me to calculate this integral.

$\endgroup$
  • $\begingroup$ It looks simpler if you push down the $\cos^2(x)$ in the denominator. You get $1$ over $\tan^2(x)+4\tan(x)$. $\endgroup$ – Nikolaj-K Dec 9 '13 at 15:09
8
$\begingroup$

HINT:

Diving the numerator & the denominator by $\sec^4x,$ $$I=\int{\dfrac{\cos^2 x}{\sin^2 x + 4\sin x \cos x}} dx=\int\frac{\sec^2xdx}{(\tan^2x+4\tan x)(\tan^2x+1)}$$

Setting $\tan x=u,$

$$I=\int\frac{du}{(u^2+4u)(u^2+1)}$$

Using Partial Fraction Decomposition

$$\frac1{u(u+4)(u^2+1)}=\frac Au+\frac B{u+4}+\frac{Cu+D}{u^2+1}$$

$\endgroup$

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.