Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$

I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.

Any hints on how to do it please?

  • $\begingroup$ The second formula does not seems to be correct. But the idea would be the tangent half-angle substitution $t=\tan(x)$. Then partial fraction decomposition would lead to the result. $\endgroup$ – Claude Leibovici Apr 24 '15 at 15:28
  • $\begingroup$ @ClaudeLeibovici don't we use $\tan x$ (instead of $\frac x 2$) only when all the powers of sin and cos are even? $\endgroup$ – shinzou Apr 24 '15 at 15:29

The denominator of your second integral should be $-3+5\cos 2x$, because from the identities $\sin x\cos x=\frac{\sin 2x}{2}$ and $\sin ^{2}x=\frac{ 1-\cos 2x}{2}$ we obtain

\begin{equation*} \frac{1+\sin x\cos x}{1-5\sin ^{2}x}=\frac{2+\sin 2x}{-3+5\cos 2x}. \end{equation*}

To evaluate

\begin{equation*} \int \frac{2+\sin 2x}{-3+5\cos 2x}dx \end{equation*}

we can use the standard half-angle substitution $t=\tan x$. Since $dt=\left( t^{2}+1\right) dx$, $\sin 2x=\frac{2t}{t^{2}+1}$ and $\cos 2x=\frac{1-t^{2}}{ t^{2}+1}$, we thus have

\begin{eqnarray*}\int \frac{2+\sin 2x}{-3+5\cos 2x}dx =\int \frac{t^{2}+t+1}{-\left( 2t-1\right) \left( 2t+1\right) \left( t^{2}+1\right) }dt \end{eqnarray*}

Now, using partial fractions decomposition, we obtain

\begin{equation*} \frac{t^{2}+t+1}{-\left( 2t-1\right) \left( 2t+1\right) \left( t^{2}+1\right) }=-\frac{7/20}{t-1/2}+\frac{3/20}{t+1/2}+\frac{t/5}{t^{2}+1}. \end{equation*}

As such,

\begin{eqnarray*} \int \frac{2+\sin 2x}{-3+5\cos 2x}dx &=&-\frac{7}{20}\ln \left| \tan x-\frac{1}{2}\right| +\frac{3}{20}\ln \left| \tan x+\frac{1}{2}\right| \\&&+\frac{1}{10}\ln \left| \tan ^{2}x+1\right| +C. \end{eqnarray*}


We have $$\int\frac{1+\sin x\cos x}{1-5\sin^2x}\mathrm{d}x =\int\frac{\mathrm{d}x}{1-5\sin^2x}+\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x$$ and $$\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x =-\frac{1}{10}\int\frac{\mathrm{d}\left(1-5\sin^2x\right)}{1-5\sin^2x} =-\frac{1}{10}\ln\left(1-5\sin^2x\right).$$ Then, $$\int\frac{\mathrm{d}x}{1-5\sin^2x} =\int\frac{\mathrm{d}x}{\cos^2x-4\sin^2x} =\int\frac{1}{1-4\tan^2x}\frac{\mathrm{d}x}{\cos^2x} =\frac{1}{2}\int\frac{\mathrm{d}\left(2\tan x\right)}{1-\left(2\tan x\right)^2} =\frac{1}{2}\tanh^{-1}\left(2\tan x\right).$$

  • $\begingroup$ What does the $d(...)$ mean? is it $dx$? $\endgroup$ – shinzou Apr 24 '15 at 15:34
  • $\begingroup$ It is the differentiation operation : for example, $\mathrm{d}\left(1-5\sin^2x\right)=-10\sin x\cos x\mathrm{d}x$. $\endgroup$ – Nicolas Apr 24 '15 at 15:36
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    $\begingroup$ The final result in the form $\frac12\ln\Bigl(\dfrac{1+2\tan x}{1-2\tan x}\Bigr)$ may be more expressive. $\endgroup$ – Bernard Apr 24 '15 at 15:40
  • $\begingroup$ I did not try to simplify it, but your formula is indeed very nice, well done! $\endgroup$ – Nicolas Apr 24 '15 at 15:41
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    $\begingroup$ No, I meant $$\tanh^{-1}\left(y\right)=\frac{1}{2}\ln\left(\frac{1+y}{1-y}\right)$$ for all $|y|<1$. It is a general formula. $\endgroup$ – Nicolas Apr 24 '15 at 16:18

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