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Solution
Your input: calculate $$$\int_{\frac{\pi}{2}}^{0}\left( \frac{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}{3} \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{\frac{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}{3} d x}=- \frac{\cos{\left(x \right)}}{12} - \frac{\cos{\left(3 x \right)}}{36}$$$ (for steps, see indefinite integral calculator)
According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.
$$$\left(- \frac{\cos{\left(x \right)}}{12} - \frac{\cos{\left(3 x \right)}}{36}\right)|_{\left(x=0\right)}=- \frac{1}{9}$$$
$$$\left(- \frac{\cos{\left(x \right)}}{12} - \frac{\cos{\left(3 x \right)}}{36}\right)|_{\left(x=\frac{\pi}{2}\right)}=0$$$
$$$\int_{\frac{\pi}{2}}^{0}\left( \frac{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}{3} \right)dx=\left(- \frac{\cos{\left(x \right)}}{12} - \frac{\cos{\left(3 x \right)}}{36}\right)|_{\left(x=0\right)}-\left(- \frac{\cos{\left(x \right)}}{12} - \frac{\cos{\left(3 x \right)}}{36}\right)|_{\left(x=\frac{\pi}{2}\right)}=- \frac{1}{9}$$$
Answer: $$$\int_{\frac{\pi}{2}}^{0}\left( \frac{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}{3} \right)dx=- \frac{1}{9}\approx -0.111111111111111$$$