evaluation of $\int\frac{1}{\sin^3 x-\cos^3 x}dx$

Solution 1:

You're missing a minus sign at one point, but other than that I think you're OK. Next, use partial fractions: $$ \frac{1}{(2-t^2)(3-t^2)} = \frac{A}{\sqrt{2}-t} + \frac{B}{\sqrt{2}+t} + \frac{C}{\sqrt{3}-t} + \frac{D}{\sqrt{3}+t} $$

Solution 2:

I had just learnt an useful technique to deal with $1+\sin x \cos x$ and want to share with you now. $$ \begin{array}{l} \displaystyle \because \frac{3}{\sin ^{3} x-\cos ^{3} x}=\frac{2}{\sin x-\cos x}+\frac{\sin x-\cos x}{1+\sin x \cos x}\\ \displaystyle \therefore \int \frac{d x}{\sin ^{3} x-\cos ^{3} x}\\ \displaystyle =\frac{2}{3} \int \frac{d x}{\sin x-\cos x}+\frac{1}{3} \int \frac{\sin x-\cos x}{1+\sin x \cos x} d x\\ \displaystyle =-\frac{2}{3 \sqrt{2}} \int \frac{d x}{\cos \left(x+\frac{\pi}{4}\right)}-\frac{2}{3} \int \frac{d(\sin x+\cos x)}{1+(\sin x+\cos x)^{2}}\\ \displaystyle =-\frac{\sqrt{2}}{3} \ln \left|\sec \left(x+\frac{\pi}{4}\right)+\tan \left(x+\frac{\pi}{4}\right)\right|-\frac{2}{3} \tan ^{-1}(\sin x+\cos x)+C\\ \displaystyle =-\frac{\sqrt{2}}{3} \ln \left|\frac{1+\frac{1}{\sqrt{2}}(\sin x+\cos x)}{\frac{1}{\sqrt{2}}(\cos x-\sin x)}\right|-\frac{2}{3} \tan ^{-1}(\sin x+\cos x)+C\\ \displaystyle =-\frac{\sqrt{2}}{3} \ln \left|\frac{\sqrt{2}+\sin x+\cos x}{\cos x-\sin x}\right|-\frac{2}{3} \tan ^{-1}(\sin x+\cos x)+C \end{array} $$ By the way, replacing $x$ by $-x$ yields its partner integral $$ \int \frac{d x}{\sin ^{3} x+\cos ^{3} x}=-\frac{\sqrt{2}}{3} \ln \left|\frac{\sqrt{2}-\sin x+\cos x}{\cos x+\sin x}\right|+\frac{2}{3} \tan ^{-1}(\sin x-\cos x)+C $$