Integrating $\int {\frac{sin^{2}x}{1+sin^{2}x}dx}$ Solution

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SUMMARY

The integral $\int {\frac{sin^{2}x}{1+sin^{2}x}dx}$ can be solved using the substitution \( t = \tan \frac{x}{2} \), leading to the transformed integral \( \int {\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt} \). After applying partial fractions, the integral separates into two parts: \( \int {\frac{2}{1+t^{2}} dt} \) and \( \int {\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt} \). The first part simplifies to \( 2 \tan^{-1}(t) \), while the second requires further analysis to identify its integral. The discussion confirms that partial fractions are essential for progressing through the solution.

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Homework Statement



\int {\frac{sin^{2}x}{1+sin^{2}x}dx}

Homework Equations



Let t = tan x/2, then dx = 2/(1+t^2) and sin x = 2t / (1+t^2)

The Attempt at a Solution



I got up to the point where \int {\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt}. Not sure if I'm on the right track and if I am, do I use partial fractions after this?

The final answer is attached. Can't really make out the handwriting :/
 

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Yes, it looks like partial fractions is the way to go after your substitution.
 
Hmm after I do partial fractions, I get
\int {\frac{2}{1+t^{2}} + {\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt}

After this, I do not know what's the next step. Kindly advise. Thanks.
 
you are summing 2 functions of t , one of these two look very much like a derivative of a certain function..
 
If you mean 2tan^-1 t, I can get this part. But what about the 2nd function?
 

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