Resolved: "Evaluate the integrals ∫0πsin25rdr"

floymdiT

Answered question

2021-10-16

Evaluate the integrals

\int_{0}^{π} \sin^{2} 5 r d r

Nathanael Webber

Skilled2021-10-17Added 117 answers

The given integral is

\int_{0}^{π} \sin^{2} 5 r d r

formula used:

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

\Rightarrow \sin^{2} t = \frac{1 - \cos (2 t)}{2}

\int_{0}^{π} \sin^{2} 5 r d r = \int_{0}^{π} \frac{1}{2} (1 - \cos (10 r)) d r

= \frac{1}{2} \int_{0}^{π} (1 - \cos (10 r)) d r

= \frac{1}{2} [\int_{0}^{π} (1) d r - \int_{0}^{π} (\cos (10 r)) d r]

= \frac{1}{2} ({[r]}_{0}^{π} - {[\frac{1}{10} \sin (10 r)]}_{0}^{π})

= \frac{1}{2} ([π - 0] - [\frac{1}{10} \sin (10 π) - \frac{1}{10} \sin (10 \cdot 0)])

= \frac{1}{2} (π - 0)

= \frac{π}{2}

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