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

floymdiT 2021-10-16 Answered
Evaluate the integrals
0πsin25rdr
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Expert Answer

Nathanael Webber
Answered 2021-10-17 Author has 117 answers
The given integral is 0πsin25rdr
formula used:
cos2t=12sin2t
sin2t=1cos(2t)2
0πsin25rdr=0π12(1cos(10r))dr
=120π(1cos(10r))dr
=12[0π(1)dr0π(cos(10r))dr]
=12([r]0π[110sin(10r)]0π)
=12([π0][110sin(10π)110sin(100)])
=12(π0)
=π2
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