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$$ \int {dx \over {\sin^3 x+\cos^3 x}}$$

Can this integral be found by substitution or any other method such as complex number?

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Since it is an indefinite integral, the solution would have to be a function, not a number. And “this substitution” that is not working, are you referring to a comment that is now removed? The standard substitution for these cases is $u=\tan(x/2)$, by the way. – Harald Hanche-Olsen Jun 3 '13 at 17:55
First cheating^3+x%2Bcos^3+x%29+dx – Ma Ming Jun 3 '13 at 17:59
Are you aware of the Weierstraß substitution? – joriki Jun 3 '13 at 18:00
@joriki I admit I wasn't aware of that name for the substitution. – Harald Hanche-Olsen Jun 3 '13 at 18:04
@Harald: I admit that I'd overlooked that you'd already suggested the substitution without mentioning the name :-) – joriki Jun 3 '13 at 19:14
up vote 6 down vote accepted

$$I=\int \frac{dx}{\sin^3x+\cos^3x}=\int \frac{dx} {\left(\sin(x)+\cos(x)\right)\left(1-\sin(x)\cos(x)\right)}$$ Write $$\sin(x)\cos(x)= \frac{\left(1-(\sin(x)-\cos(x))^2\right)}{2}$$ So

$$ I=\int \frac{dx} {\left(\sin(x)+\cos(x)\right) \left(1-\frac{\left(1-(\sin(x)-\cos(x))^2\right)}{2}\right)}$$ Now Use Substitution $$\sin(x)-\cos(x)=t$$

$$ \left(\cos(x)+\sin(x)\right)dx=dt \implies dx=\frac{dt}{\left(\sin(x)+\cos(x)\right)}$$ So

$$ \frac{dx}{\left(\cos(x)+\sin(x)\right)}=\frac{dt}{\left(\sin(x)+\cos(x)\right)^2} =\frac{dt}{1+2\sin(x)\cos(x)}=\frac{dt}{1+2\frac{\left(1-(\sin(x)-\cos(x))^2\right)}{2}}=\frac{dt}{2-t^2}$$

So $$I=\int \frac{2dt}{\left(t^2+1\right)\left(2-t^2\right)}=\frac{2}{3}\left( \int \frac{dt}{t^2+1}-\int \frac{dt}{t^2-2}\right)$$ So

$$ I=\frac{2}{3} \tan^{-1}(t)-\frac{1}{3\sqrt{2}}\ln\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|$$

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You can apply \left and \right to | just like you apply it to parentheses and the like. – joriki Jun 3 '13 at 19:15
Yeah sure, Thank you – Ekaveera Kumar Sharma Jun 4 '13 at 2:01

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