# Evaluate the integrals \int_0^\pi\sin^2 5rdr

Evaluate the integrals
${\int }_{0}^{\pi }{\mathrm{sin}}^{2}5rdr$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Nathanael Webber
The given integral is ${\int }_{0}^{\pi }{\mathrm{sin}}^{2}5rdr$
formula used:
$\mathrm{cos}2t=1-2{\mathrm{sin}}^{2}t$
$⇒{\mathrm{sin}}^{2}t=\frac{1-\mathrm{cos}\left(2t\right)}{2}$
${\int }_{0}^{\pi }{\mathrm{sin}}^{2}5rdr={\int }_{0}^{\pi }\frac{1}{2}\left(1-\mathrm{cos}\left(10r\right)\right)dr$
$=\frac{1}{2}{\int }_{0}^{\pi }\left(1-\mathrm{cos}\left(10r\right)\right)dr$
$=\frac{1}{2}\left[{\int }_{0}^{\pi }\left(1\right)dr-{\int }_{0}^{\pi }\left(\mathrm{cos}\left(10r\right)\right)dr\right]$
$=\frac{1}{2}\left({\left[r\right]}_{0}^{\pi }-{\left[\frac{1}{10}\mathrm{sin}\left(10r\right)\right]}_{0}^{\pi }\right)$
$=\frac{1}{2}\left(\left[\pi -0\right]-\left[\frac{1}{10}\mathrm{sin}\left(10\pi \right)-\frac{1}{10}\mathrm{sin}\left(10\cdot 0\right)\right]\right)$
$=\frac{1}{2}\left(\pi -0\right)$
$=\frac{\pi }{2}$