Question

Evaluate ${\int }_0^{\pi /2}\frac{{\cos }^{9}x}{{\cos }^{3}x+{\sin }^{3}x}dx$.

Answer 1

$\begin{array}{c}I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{9}x}{{\cos }^{3}x+{\sin }^{3}x}dx\to \left(i\right)\\ I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^9\left(\frac{\pi }{2}-x\right)}{{\cos }^3\left(\frac{\pi }{2}-x\right)+{\sin }^3\left(\frac{\pi }{2}-x\right)}\\ I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^9\left(x\right)}{{\sin }^3\left(x\right)+{\cos }^3\left(x\right)}dx\end{array}$

adding (i) & (ii)

$\begin{array}{c}2I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{9}x+{\sin }^{9}x}{{\cos }^{3}x+{\sin }^{3}x}dx\\ 2I={\int }_0^{\frac{\pi }{2}}\frac{{\left({\cos }^{3}x\right)}^3+{\left({\sin }^{3}x\right)}^3}{{\sin }^{3}x+{\cos }^{3}x}dx\\ 2I=\int \frac{\left({\cos }^{3}x+{\sin }^{3}x\right)+{\left({\cos }^{3}x\right)}^2+\left({\sin }^{3}x\right)-{\sin }^{3}x.{\cos }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx\\ 2I={\int }_0^{\frac{\pi }{2}}{\left({\cos }^{3}x\right)}^2+{\left({\sin }^{3}x\right)}^2-{\sin }^{3}x.{\cos }^{3}xdx\\ ={\int }_0^{\frac{\pi }{2}}{\cos }^{3}x\left\{{\cos }^{3}x-{\sin }^{3}x\right\}+{\sin }^{3}x\left\{{\sin }^{3}x-{\cos }^{3}x\right\}+\left\{{\sin }^{3}x+{\cos }^{3}x\right\}dx\\ ={\int }_0^{\frac{\pi }{2}}{\cos }^{3}x\left\{\left(\cos x+\sin x\right)\left\{1+\cos x.\sin x\right\}\right\}\end{array}$