# Computing $\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)}$

$$\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)}$$ This question was asked today in my maths exam, It was one of those two questions which I couldn't answer, How do you go about answering it ?

• I think the answer below will clear your doubts. Sorry didn't look before. Oct 14, 2015 at 13:41

\begin{align}\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} dx &= \int \frac{4\tan(x)+5}{\cos^2(x)(\tan (x)+1) (\tan (x)+2)}dx\\ &= \int \frac{4u+5}{(u+1) (u+2)}du \end{align} The last part is by using $u=\tan x$, $du = \frac{dx}{\cos^2 x}$. I hope you know to continue from here.

Note: $$\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x) =\cos^2(x)( \tan^2(x)+2+3\tan(x))$$ denote $\tan x = u$ to get $\tan^2(x)+2+3\tan(x) = u^2+2+3u = (u+1)(u+2)$

• Not clear about the denominator simplification Oct 14, 2015 at 13:39
• To me it is quite clear.
– JSCB
Oct 14, 2015 at 13:42
• Can you tell me how he simplified it to that? Oct 14, 2015 at 13:44
• Bear in mind that $\tan x=\sin x/\cos x$, expand the brackets.
– JSCB
Oct 14, 2015 at 13:45
• @CuriousSciDude I've added the denominator simplification for you. Oct 14, 2015 at 13:57