Showing that \int_0^{ \frac \pi 2} \cos^{2n+1}(x) dx =

luminarc24lry

luminarc24lry

Answered question

2022-04-30

Showing that 0π2cos2n+1(x)dx=4n(n!)2(2n+1)!?

Answer & Explanation

Brenton Steele

Brenton Steele

Beginner2022-05-01Added 18 answers

We have
0π2cos2n+1(x)dx
=0π2cos(x)cos2n(x)dx
=[sin(x)cos2n(x)]0π2+0π22nsin2(x)cos2n1(x)dx
=2n0π2(1cos2x)cos2n1(x)dx
=2n(0π2cos2n1(x)dx0π2cos2n+1(x)dx)
and therefore
0π2cos2n+1(x)dx=2n2n+10π2cos2n1(x)dx
Besides, 0π2cos(x)dx=1. So
0π2cos2n+1(x)dx=2n2n+1×2n22n1××1

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