MHB Integrating Trigonometric Functions with Powers and Substitution

[karush]

$\large{S6.7.R.27}$
$$\displaystyle
I=\int_{0}^{\pi/2}\cos^3\left({x}\right)\sin\left({2x}\right) \,dx = \frac{2}{5}$$
since $\sin\left({2x}\right)=2\sin\left({x}\right)\cos\left({x}\right) $ then..
$$I=2\int_{0}^{\pi/2}\cos^4\left({x}\right)\sin\left({x}\right)\, dx \\
\begin{align}
u&=\cos\left({x}\right) &du=-\sin\left({x}\right) \, dx
\end{align}$$
then
$$\displaystyle
I=\int_{0}^{\pi/2}{u}^{4} \, du$$

I continued but didn't get the answer for some reason

 

[Theia]

When you do a substitution, what happen to the limits of the integral?

 

[karush]

$\large{S6.7.R.27}$
$$\displaystyle
I=\int_{0}^{\pi/2}\cos^3\left({x}\right)\sin\left({2x}\right) \,dx = \frac{2}{5}$$
since $\sin\left({2x}\right)=2\sin\left({x}\right)\cos\left({x}\right) $ then..
$$I=-2\int_{0}^{\pi/2}\cos^4\left({x}\right)\sin\left({x}\right)\, dx \\
\begin{align}
u&=\cos\left({x}\right) &du&=-\sin\left({x}\right) \, dx \\
0&=\cos\left({\frac{\pi}{2}}\right) &1 &=\cos\left({0}\right)
\end{align}$$
then
$$\displaystyle
I=2\int_{0}^{1} {u}^{4} \, du
=2\left| \frac{{u}^{5}}{5} \right|_{0}^1$$

 

[Greg]

$\cos(0)=1$, so you've got

$$-2\int_1^0u^4\,du=2\int_0^1u^4\,du$$