Question

Solve:

$\int \frac{{\sin }^{4}x+{\cos }^{4}x}{{\sin }^{3}x+{\cos }^{3}x}$

Answer 1

$I=\int \frac{{\sin }^{4}x+{\cos }^{4}x}{{\sin }^{3}x+{\cos }^{3}x}dx$

$=\int [(\cos x+\sin x)-\frac{({\cos }^{2}x+{\sin }^{2}x)\sin x\cos x}{(\sin x+\cos x)({\sin }^{2}x-\sin x\cos x+{\cos }^{2}x)}]dx$

$=\int (\cos x+\sin x)dx-\int \frac{({\cos }^{2}x+{\sin }^{2}x)\sin x\cos x}{(\sin x+\cos x)({\sin }^{2}x-\sin x\cos x+{\cos }^{2}x)}dx$

$I^{st}$ part

$I{I}^{nd}$ part

Integrating II part

$I_2=\int \frac{-\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}$

Now let $u=\sin x-\cos x$

$du=(\cos x+\sin x)dx$

$u^2=1-2\sin x\cos x$

$\Rightarrow \sin x\cos x=(1-u^2)/2$

$\therefore I_2=\int \frac{-(1-u^2)/2}{(\sin x+\cos x{)}^2(1-\frac{1-u^2}{2})}du$

$I=\int \frac{1-u^2}{(2-u^2)(1+u^2)}du$

$=\frac{1}{6\sqrt{2}}log\left|\frac{\sqrt{2}+u}{\sqrt{2}-u}\right|-\frac{2}{3}{\tan }^{-1}u+c_1$

$=\frac{1}{6\sqrt{2}}log\left|\frac{\sqrt{2}+\sin x-\cos x}{\sqrt{2}-\sin x+\cos x}\right|-\frac{2}{3}{\tan }^{-1}(\sin x-\cos x)+c_1$

And $I_1=\int \cos xdx+\int \sin xdx=\sin x-\cos x+c_2$

$\therefore I=I_1+I_2=\sin x-\cos x+\frac{1}{6\sqrt{2}}log\left|\frac{\sqrt{2}+\sin x-\cos x}{\sqrt{2}-\sin x+\cos x}\right|-\frac{2}{3}{\tan }^{-1}(\sin x-\cos x)+c$

where $c=c_1+c_2$.