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Try an interesting integration problem:

$$\int\frac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}{\rm d}x$$

I know the solution which almost took me 15 minutes. Good Luck!

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closed as off-topic by Carl Mummert, Davide Giraudo, graydad, Ivo Terek, anomaly Mar 8 '15 at 18:49

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  • 2
    $\begingroup$ This solution took me only $2$ minutes. $\endgroup$ – Elaqqad Mar 5 '15 at 13:55
  • $\begingroup$ @Elaqqad check the link again. $\endgroup$ – RE60K Mar 5 '15 at 13:58
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    $\begingroup$ It's not a good idea (using always mathematica) but it's the first thing i do when i was given a hazardous calculus, I'm an addict! so much respect for you you don't use them. $\endgroup$ – Elaqqad Mar 5 '15 at 14:07
  • $\begingroup$ @Elaqqad it sometimes gives wierd forms. In this case too. $\endgroup$ – RE60K Mar 5 '15 at 14:09
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take the $\cos^3x$ common from denominator outside , you will get $$\frac{\tan^2x\sec^2x}{(\tan^3x+1)^2}$$ now take $t=\tan^3x+1$ thus $dt=3\tan^2x\sec^2x$ thus easily solvable

$$\int\frac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}{\rm d}x=-\frac13\frac1{1+\tan^3x}+c$$

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    $\begingroup$ Yes correct....Bravo $\endgroup$ – RE60K Mar 5 '15 at 13:59
  • $\begingroup$ @avz2611 Did you do the "trick" at first glance? It always takes me a lot of time dealing with such problems. $\endgroup$ – Ivy Mar 5 '15 at 14:43
  • $\begingroup$ not at the first glance , i had a different idea the first time , but i did get it at the second attempt, these kind of problems might be first trick for one but take 15 mins for other but it can easily be vice versa $\endgroup$ – avz2611 Mar 5 '15 at 17:01
  • $\begingroup$ this is a good answer to an interesting question. To keep it here, someone should add a post to the Reopening Thread lest it be deleted. $\endgroup$ – robjohn Mar 12 '15 at 15:21

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