1. ## 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

2. 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.

3. Just use the vastly underrated method of Weierstrass Substitution... then both integrals are easy as pie...

4. 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 !

5. 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 (;->)