Crazy integral

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July 9th 2009, 12:10 AM
simplependulum

Crazy integral

Determine which integral is more crazy ...

$\int \frac{dx}{1 + \sin^3{x}}$

and

$\int \frac{dx}{1+\sin^6{x}}$

3 or 6 ? The answer is 3 ! It is very interesting that although 6 is greater than 3 , the integral $\int \frac{dx}{1+\sin^6{x}}$ is actually easier than the other .

I suggest to try $\int \frac{dx}{1+\sin^6{x}}$ first , finally finish the rest (Evilgrin)
July 9th 2009, 06:26 PM
halbard

If I were to attempt to find $\int\frac1{1+\sin^3 x}\mathrm dx$ then I'd be crazy. I have attempted to find it, and I am crazy ... now.

Thanks a lot, simplependulum! You have determined which member is more crazy.
July 9th 2009, 07:48 PM
AlephZero

Just use the vastly underrated method of Weierstrass Substitution... then both integrals are easy as pie...
July 9th 2009, 08:46 PM
simplependulum

Quote:

Originally Posted by halbard

If I were to attempt to find $\int\frac1{1+\sin^3 x}\mathrm dx$ then I'd be crazy. I have attempted to find it, and I am crazy ... now.

Thanks a lot, simplependulum! You have determined which member is more crazy.

Yes , people can prove they are crazy or not in this question .

I was wondering which method you have made use , I first deduced a general solution for $\int \frac{ dx}{ \sin^2{x} + b \sin{x} + c } ~~$ and if the module of the root $u^2 + bu + c = 0$ is 1 , it will become easier to solve !
July 10th 2009, 02:51 AM
halbard

Are you by chance referring to the identity

$\frac3{1+\sin^3 x}=\dfrac1{1+\sin x}+\frac1{1-\omega_1\sin x}+\frac1{1-\omega_2\sin x}$

where $\omega_1$ , $\omega_2$ are the complex cube roots of $-1$ , together with the substitution $t=\tan{\textstyle\frac12}x$ for $\int\dfrac1{1-a\sin x}\mathrm dx$ which can be found when $a^2\neq 1$ using $\int\frac1{1-2at+t^2}\mathrm dt=\frac{\arctan\left(\displaystyle\frac{t-a}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}$ ? No, I haven't tried it this way (;->)