Evaluate the following integrals. \int_0^{\frac{\pi}{2}}\sin^7xdx

tinfoQ 2021-10-08 Answered
Evaluate the following integrals.
0π2sin7xdx
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Expert Answer

Ezra Herbert
Answered 2021-10-09 Author has 99 answers
Given 0π2sin7xdx
sin7(x)=sin6(x)sin(x)
and sin6(x)=(sin2(x))3
and sin2(x)=1cos2(x)
0π2sin6(x)sin(x)dx
=0π2(sin2(x))3sinxdx
=0π2(1cos2(x))3sinxdx
Apply u-substitution u=cos(x)du=sinxdx
du=sinxdx
when x=0,u=cos(0)=1
when x=π2,u=cos(π2)=0
Apply the formula (ab)3=a3b33a2b+3ab2
=01[((1)3(u2))33(1)2(u2)+3(1)(u2)2]du
=01(1u63u2+3u4)du
=(10)17(10)+35(10)
=1171+35=17+35
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