Evaluate the integrals

${\int}_{0}^{\pi}{\mathrm{sin}}^{2}5rdr$

floymdiT
2021-10-16
Answered

Evaluate the integrals

${\int}_{0}^{\pi}{\mathrm{sin}}^{2}5rdr$

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Nathanael Webber

Answered 2021-10-17
Author has **117** answers

The given integral is ${\int}_{0}^{\pi}{\mathrm{sin}}^{2}5rdr$

formula used:

$\mathrm{cos}2t=1-2{\mathrm{sin}}^{2}t$

$\Rightarrow {\mathrm{sin}}^{2}t=\frac{1-\mathrm{cos}\left(2t\right)}{2}$

${\int}_{0}^{\pi}{\mathrm{sin}}^{2}5rdr={\int}_{0}^{\pi}\frac{1}{2}(1-\mathrm{cos}\left(10r\right))dr$

$=\frac{1}{2}{\int}_{0}^{\pi}(1-\mathrm{cos}\left(10r\right))dr$

$=\frac{1}{2}[{\int}_{0}^{\pi}\left(1\right)dr-{\int}_{0}^{\pi}\left(\mathrm{cos}\left(10r\right)\right)dr]$

$=\frac{1}{2}({\left[r\right]}_{0}^{\pi}-{\left[\frac{1}{10}\mathrm{sin}\left(10r\right)\right]}_{0}^{\pi})$

$=\frac{1}{2}([\pi -0]-[\frac{1}{10}\mathrm{sin}\left(10\pi \right)-\frac{1}{10}\mathrm{sin}(10\cdot 0)])$

$=\frac{1}{2}(\pi -0)$

$=\frac{\pi}{2}$

formula used:

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