Finding $\displaystyle \int\frac{\sin^8 x+\cos^9 x}{1+\tan^9 x}dx$
$\bf{Attempt:}$ Integral $\displaystyle I = \int\frac{(\sin^8 x+\cos^9 x)}{\sin^9 x+\cos^9 x}\cdot \cos^9 xdx $
Put $(\sin^9 x+\cos^9 x) = t,$ then $9\sin x\cos x(\sin^7 x+\cos^7 x)dx = dt$
Could some help me how to solve it, Thanks
I am almost sure that there is a typo in the problem which is probably $$\int\frac{\sin^\color{red}{9} (x)+\cos^9 (x)}{1+\tan^9 (x)}\,dx=\int \cos^9 (x)\,dx $$ which is simple using linear combinations of $\cos(nx)$ when $n$ is odd $$\cos^9(x)=\frac{63 }{128}\cos (x)+\frac{21}{64} \cos (3 x)+\frac{9}{64} \cos (5 x)+\frac{9}{256} \cos (7 x)+\frac{1}{256} \cos (9 x)$$ Otherwise, as Benjamin Dickman commented, it is a pure mess !
$$ \begin{aligned} \int\frac{\sin^\color{red}{9} (x)+\cos^9 (x)}{1+\tan^9 (x)}\,dx=&\int \cos^9 x\,dx \\= & \int\left(1-\sin ^2 x\right)^4 d(\sin x) \\ = & \int\left(1-4 \sin ^2 x+6 \sin ^4 x-4 \sin ^6 x+\sin ^2 x\right) d(\sin x) \\ = & \sin x-\frac{4 \sin ^3 x}{3}+\frac{6 \sin ^5 x}{5}-\frac{4 \sin ^7 x}{7} + \frac{\sin ^2 x}{9}+C \end{aligned} $$
