How do you evaluate the integral? int_0^(pi/2)cos^(3)x sin2x dx

Oct 3, 2017

Substitute $2 \sin \left(x\right) \cos \left(x\right)$ for $\sin \left(2 x\right)$
Perform a "u" substitution and change of limits
Integrate
Evaluate at the limits:

Explanation:

Given ${\int}_{0}^{\frac{\pi}{2}} {\cos}^{3} \left(x\right) \sin \left(2 x\right) \mathrm{dx}$

Substitute $2 \sin \left(x\right) \cos \left(x\right)$ for $\sin \left(2 x\right)$

$2 {\int}_{0}^{\frac{\pi}{2}} {\cos}^{4} \left(x\right) \sin \left(x\right) \mathrm{dx}$

Let $u = \cos \left(x\right)$, then $\frac{\mathrm{du}}{\mathrm{dx}} = - \sin \left(x\right)$

A more suitable form for substitution:

$\sin \left(x\right) \mathrm{dx} = - \mathrm{du}$

$b = \cos \left(\frac{\pi}{2}\right) = 0$

$a = \cos \left(0\right) = 1$

$- 2 {\int}_{1}^{0} {u}^{4} \left(x\right) \mathrm{du}$

Negate and flip the limits:

$2 {\int}_{0}^{1} {u}^{4} \left(x\right) \mathrm{du} = \frac{2}{5} {\left({u}^{5}\right]}_{0}^{1} = \frac{2}{5} \left({1}^{5} - {0}^{5}\right)$

${\int}_{0}^{\frac{\pi}{2}} {\cos}^{3} \left(x\right) \sin \left(2 x\right) \mathrm{dx} = \frac{2}{5}$