Evaluate some definite integrals of even powers of sine

From the previous two exercises (here and here) we know

$\int \sin^2 x \, dx = \frac{1}{2}x - \frac{1}{4} \sin (2x),$

and

$\int \sin^n x \, dx = - \frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x \, dx.$

Use these results to evaluate the following:

$\displaystyle{ \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4}}$ .
$\displaystyle{ \int_0^{\frac{\pi}{2}} \sin^4 x \, dx = \frac{3}{4} \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{3 \pi}{16}}$ .
$\displaystyle{ \int_0^{\frac{\pi}{2}} \sin^6 x \, dx = \frac{5}{6} \int_0^{\frac{\pi}{2}} \sin^4 x \, dx = \frac{5 \pi}{32}}$ .

Using the first formula we have
$\begin{align*} \int_0^{\frac{\pi}{2}} \sin^2 x \, dx &= \left. \left( \frac{1}{2} x - \frac{1}{4} \sin (2x) \right) \right|_0^{\frac{\pi}{2}} \\[9pt] &= \left( \frac{\pi}{4} - 0 \right) - 0 \\[9pt] &= \frac{\pi}{4}. \end{align*}$
Using the second formula and then part (a) we have
$\begin{align*} \int_0^{\frac{\pi}{2}} \sin^4 x \, dx &= \left.- \frac{\sin^3 x \cos x}{4}\right|_0^{\frac{\pi}{2}} + \frac{3}{4} \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \\[9pt] &= 0 + \frac{3}{4} \left( \frac{\pi}{4} \right) \\[9pt] &= \frac{3 \pi}{16}. \end{align*}$
Using the second formula and then part (b) we have
$\begin{align*} \int_0^{\frac{\pi}{2}} \sin^6 x \, dx &= \left. -\frac{\sin^5 x \cos x}{6} \right|_0^{\frac{\pi}{2}} + \frac{5}{6} \int_0^{\frac{\pi}{2}} \sin^4 x \, dx \\[9pt] &= 0 + \frac{5}{6} \left( \frac{3 \pi}{16} \right) \\[9pt] &= \frac{5 \pi}{32}. \end{align*}$

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