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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2$\begingroup$ This solution took me only $2$ minutes. $\endgroup$– ElaqqadMar 5, 2015 at 13:55
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$\begingroup$ @Elaqqad check the link again. $\endgroup$– RE60KMar 5, 2015 at 13:58
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1$\begingroup$ 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. $\endgroup$– ElaqqadMar 5, 2015 at 14:07
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$\begingroup$ @Elaqqad it sometimes gives wierd forms. In this case too. $\endgroup$– RE60KMar 5, 2015 at 14:09
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1 Answer
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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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$\begingroup$ @avz2611 Did you do the "trick" at first glance? It always takes me a lot of time dealing with such problems. $\endgroup$– IvyMar 5, 2015 at 14:43
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$\begingroup$ 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 $\endgroup$– avz2611Mar 5, 2015 at 17:01
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$\begingroup$ 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, 2015 at 15:21