# Crazy integral

• Jul 9th 2009, 12:10 AM
simplependulum
Crazy integral
Determine which integral is more crazy ...

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

and

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

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

I suggest to try $\displaystyle \int \frac{dx}{1+\sin^6{x}}$ first , finally finish the rest (Evilgrin)
• Jul 9th 2009, 06:26 PM
halbard
If I were to attempt to find $\displaystyle \int\frac1{1+\sin^3 x}\mathrm dx$ 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.
• Jul 9th 2009, 07:48 PM
AlephZero
Just use the vastly underrated method of Weierstrass Substitution... then both integrals are easy as pie...
• Jul 9th 2009, 08:46 PM
simplependulum
Quote:

Originally Posted by halbard
If I were to attempt to find $\displaystyle \int\frac1{1+\sin^3 x}\mathrm dx$ 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 $\displaystyle \int \frac{ dx}{ \sin^2{x} + b \sin{x} + c } ~~$ and if the module of the root $\displaystyle u^2 + bu + c = 0$ is 1 , it will become easier to solve !
• Jul 10th 2009, 02:51 AM
halbard
Are you by chance referring to the identity

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

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