Find $$\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$$
What I tried:
$$\sin^4(x)+\cos^4(x)=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x$$
and $$\sin^3 x+\cos^3 x=(\sin x+\cos x)(1-\sin x\cos x)$$
so $$\int\frac{1-\sin^2 x\cos^2 x}{(\sin x+\cos x)(1-\sin x\cos x)}dx-\int\frac{\sin^2 x\cos^2 x}{(\sin x+\cos x)(1-\sin x\cos x)}dx$$
How do I solve it? Help me, please.
A brutal way is just to enforce the substitution $x=2\arctan t$, leading to $$ \int\frac{\sin x\cos x}{\sin^3 x+\cos^3 x}\,dx=2\int\frac{\frac{2t(1-t^2)}{(1+t^2)^3}}{\frac{(2t)^3}{(1+t^2)^3}+\frac{(1-t^2)^3}{(1+t^2)^3}}\,dt=\int\frac{4t(1-t^2)}{(2t)^3+(1-t^2)^3}\,dt $$ an ugly integral, but perfectly solvable by partial fraction decomposition. The roots of $(2t)^3+(1-t^2)^3$ can be found by solving the quadratic equations given by $$\frac{1-t^2}{2t}\in\left\{-1,\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}\right\}.$$
We have $\dfrac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}=\sin x+\cos x-\dfrac{\sin x\cos x}{\sin^3x+\cos^3x}$
– lab bhattacharjee Feb 21 '19 at 15:06