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This seems like a simple problem, but my trig manipulations are leading to a dead end.

Compute: $$\int\frac{\sin^2(x)}{1 - \tan(x)} dx$$

Working thus far:

Replace $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$

Multiply fraction by $\dfrac{\cos(x)}{\cos(x)}$ and then by $\dfrac{\cos(x)+\sin(x)}{\cos(x)+\sin(x)}$ to get:

\begin{align*} \int\frac{\sin^2(x)\cos(x)(\cos(x) + \sin(x))}{\cos^2(x) - \sin^2(x)} dx &= \int\frac{\sin^2(x)\cos^2(x) + \sin^3(x)\cos(x)}{\cos(2x)} dx \\ &= \int\frac{\sin^2(2x)}{4\cos(2x)} dx + \int\frac{\sin^3(x)\cos(x)}{\cos(2x)} dx. \end{align*}

From there I'm not sure to proceed (in particular on the first integral). The second integral should be doable if I use $\cos(2x) = 1 - 2\sin^2(x)$ and use $ u = \sin(x)$.

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  • $\begingroup$ For your first integral write $\sin ^{2} {2x}=1-\cos^ {2} {2x}$ and for the second put $\cos x =t$. $\endgroup$ – Apurv Apr 4 '14 at 18:29
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Let $t=\tan x$ then $dt=(1+t^2)dx$ and recall that $$\sin^2 x=1-\cos^2x=1-\frac1{1+\tan^2x}=\frac{t^2}{1+t^2}$$ hence the integral becomes $$\int\frac{t^2}{(1+t^2)^2(1-t)}dt$$ can you take it from here?

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  • $\begingroup$ That's a nice method. The square in $(1 + t^2)^2$ is a typo correct? Should just be $$\frac{t^2}{(1 + t^2)(1 - t)}$$. $\endgroup$ – Grid Apr 4 '14 at 18:36
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    $\begingroup$ No it isn't a typo and you should now use the partial fraction decomposition. $\endgroup$ – user63181 Apr 4 '14 at 18:38
  • $\begingroup$ Right, sorry. Forgot the extra $(1 + t^2)$ term came from the $dt = (1 + t^2) dx$. Thanks! The rest should be doable. $\endgroup$ – Grid Apr 4 '14 at 18:42
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Apr 4 '14 at 18:45
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Let $I = \int \dfrac{sin^2x}{1 - tanx} dx$, and $J = \int \dfrac{cos^2x}{1 - tanx} dx$.

Then $I + J = \int \dfrac{1}{1 - tanx} dx$, and $J - I = \int \dfrac{cos^2x - sin^2x}{1 - tanx} dx = \int (cos^2x + sinx\cdot cosx)dx $. By solving these simpler integrals we can then find $I$ ( and $J$).

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