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As the title says, I want to know how to solve this integral: $$\int \frac{\sin x}{\tan^2x+\cos^2x}\,dx.$$ I tried to write everything as $\cos x$ and I got that the integral is equal to: $$ \int \frac{\sin x\cos^2x}{\cos^4x-\cos^2x+1}\,dx.$$ I tried to substitute $\cos x=t \implies -\sin x \,dx=dt$ and I got the integral: $$-\int \dfrac{t^2}{t^4-t^2+1}\,dt,$$ which I do not know how to solve. Is everything I have done so far right or I can do it in another way that is faster or easier?

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  • $\begingroup$ Can you factor the denominator and do a partial fraction decomposition? $\endgroup$ Nov 5 '17 at 16:40
  • $\begingroup$ Everything seems fine. As for the integral that you reach, are you aware about integration by partial fractions? $\endgroup$ Nov 5 '17 at 16:41
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    $\begingroup$ "Solve" is the wrong word. One solves problems; one solves equations; one evaluates expressions. The question is how to evaluate this integral. $\endgroup$ Nov 8 '17 at 16:45
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You did well. For the rest, use the decomposition:

\begin{align} t^4-t^2+1 & = t^4+2t^2+1-3t^2 \\ & = (t^2+1)^2-3t^2 \\ & = (t^2+\sqrt{3}t+1)(t^2-\sqrt{3}t+1). \end{align}

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