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

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

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$$\begin{align}\int \frac{4\tan(x)+5}{\sin^2(x)+2\cos^2(x)+3\sin(x)\cos(x)} dx &= \int \frac{4\tan(x)+5}{\cos^2(x)(\tan (x)+1) (\tan (x)+2)}dx\\ &= \int \frac{4u+5}{(u+1) (u+2)}du \end{align}$$ 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 $\tan^2(x)+2+3\tan(x) = u^2+2+3u = (u+1)(u+2)$

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

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