That's decided a similar example. This I do not know how to solve. Help me please.enter image description here

My integral: $$\int \frac{\left(sin\left(2 \:x\right)\right)^2}{\left(sin^3x+cos^3x\right)^2}$$

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    $\begingroup$ try the tan half angle substituion $\endgroup$ – Dr. Sonnhard Graubner Apr 8 '15 at 21:00

$\displaystyle\int\frac{(\sin 2x)^2}{(\sin^{3}x+\cos^{3}x)^2}dx=\int\frac{4\sin^2x\cos^2x}{(\sin^{3}x+\cos^{3}x)^2}dx=\int\frac{4\tan^2x\sec^2x}{(\tan^3x+1)^2}dx$

$\hspace{1.8 in}$(after dividing by $\cos^6x$ on the top and bottom).

Now let $u=\tan^3x+1, du=3\tan^2x\sec^2x\;dx$ to get


  • $\begingroup$ wow. how do you come up with this?(+1) too bad, i can upvote only once. $\endgroup$ – abel Apr 9 '15 at 0:22
  • $\begingroup$ Thanks - I just tried the techniques I've seen people like you use. $\endgroup$ – user84413 Apr 9 '15 at 0:23

hint: $$\sin(x)^3+\cos(x)^3=(\sin(x)+\cos(x))(1-\sin(x)\cos(x))$$ and $$\sin(2x)=2\sin(x)\cos(x)$$ i have got this integral $$\int\,{\frac { \left( {t}^{2}+1 \right) \left( t-1 \right) ^{2} \left( t+1 \right) ^{2}{t}^{2}}{ \left( {t}^{4}+2\,{t}^{3}+2\,{t}^{2}-2\,t+1 \right) ^{2} \left( {t}^{2}-2\,t-1 \right) ^{2}}} dt$$

  • $\begingroup$ So I got, what's next? $$\int \frac{(2sinx \cdot cosx)^{2}}{((sinx+cosx)(1-sinx \cdot cosx))^{2}}dx$$ $\endgroup$ – andre1 Apr 8 '15 at 21:12
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    $\begingroup$ use the tan half angle substitution and $\sin(x)$ looks better than $sin(x)$ $\endgroup$ – Dr. Sonnhard Graubner Apr 8 '15 at 21:14
  • $\begingroup$ I do not understand what to use. $\endgroup$ – andre1 Apr 8 '15 at 21:18
  • $\begingroup$ one moment please i will send you a link $\endgroup$ – Dr. Sonnhard Graubner Apr 8 '15 at 21:20
  • $\begingroup$ $\sin(x)=\frac{2t}{1+t^2}$,$\cos(x)=\frac{1-t^2}{1+t^2}$ $\endgroup$ – Dr. Sonnhard Graubner Apr 8 '15 at 21:22

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