Finding $\displaystyle \int\frac{1}{(\sin^3(x)+\cos^3(x))^2}dx$
Try: $$I=\int\frac{1}{(\sin^3 x+\cos^3 x)^2}dx=\int\frac{\sec^6 x}{(1+\tan^3 x)^2}dx$$
Substituting $\tan x=t$ and $\sec^2 xdx=dt$
$$I=\int\frac{(1+t^2)^2}{(1+t^3)^2}dt$$
Could some help me to solve it , thanks
You could try a partial fraction decomposition for the integrand. Here $$\frac{(1 + t^2)^2}{(1 + t^3)^2} = \frac{t}{3(t^2 - t + 1)^2} + \frac{5}{9(t^2 - t + 1)} + \frac{4}{9(t + 1)^2},$$ and integrate each of these rational functions separately.
$$\begin{align}\int\frac{t^4}{(1+t^3)^2}dt&=\frac{-\frac13t^2}{(1+t^3)}+\frac23\int\frac t{1+t^3}dt\\ \int\frac{2t^2}{(1+t^3)^2}dt&=\frac{-\frac23}{1+t^3}+C_1\\ \int\frac1{(1+t^3)^2}dt&=\int\frac{1+t^3}{(1+t^3)^2}dt-\int\frac{t^3}{(1+t^3)^2}dt\\ &=\int\frac1{1+t^3}dt-\left\{\frac{-\frac13t}{1+t^3}+\frac13\int\frac1{1+t^3}dt\right\}\end{align}$$ So $$\begin{align}\int\frac{t^4+2t^2+1}{(1+t^3)^2}dt&=\frac{-\frac13t^2+\frac13t-\frac23}{1+t^3}+\int\frac{\frac23t+1-\frac13}{1+t^3}dt\\ &=\frac{-\frac13t^2+\frac13t-\frac23}{1+t^3}+\frac23\int\frac1{\left(t-\frac12\right)^2+\left(\frac{\sqrt3}2\right)^2}dt\\ &=\frac{-\frac13t^2+\frac13t-\frac23}{1+t^3}+\frac23\frac2{\sqrt3}\tan^{-1}\left(\frac{t-\frac12}{\frac{\sqrt3}2}\right)+C\end{align}$$
