I am trying to find this by using integration by

Aine Sellers

Aine Sellers

Answered question

2022-02-25

I am trying to find this by using integration by parts but I am not sure how to do it.
0π2(sinx)7(cosx)5dx

Answer & Explanation

nastaja1en

nastaja1en

Beginner2022-02-26Added 2 answers

To integrate the function
f(x)=sinnxcosmx,
when n or m are positive odd numbers, we can apply a general technique which consists of expanding f(x) into a sum of terms of the form
sinpxcosx, p=1,2,
or
cosqxsinx, q=1,2,
Using the identity
cos2x=1sin2x
in the form
cos4x=(1sin2x)2=12sin2x+sin4x,
we rewrite our
f(x)=sin7xcos5x=sin7xcos4xcosx as
as
f(x)=sin7x(12sin2x+sin4x)cosx
=sin7xcosx2sin9xcosx+sin11xcosx
Each term is of the form sinpxcosx and can easily be integrated by the substitution u=sinx, u=cosx, du=cosxdx=udx:
sinpxcosxdx=updu=up+1p+1=sinp+1xp+1+C
0π2sinpxcosxdx=1p+1
0π2f(x)dx=0π2sin7xcos5xdx
shotokan0758s

shotokan0758s

Beginner2022-02-27Added 8 answers

Let I=0π2(sinx)7(cosx)5dx
Then using the substitution u=π2x we have I=0π2(sinx)5(cosx)7dx
Adding gives
2I=0π2(sinx)5(cosx)5dx
Since cos2(x)+sin2(x)=1
And that
2I=1250π2(sin2x)5dx
since sin(2x)=2sin(x)cos(x). Then substituting u=2x gives
2I=1260π(sinu)5du=1260π(sinu)(1cos2u)2du
=1260π(sinu)(12cos2u+cos4(u))du
Hence
I=127[cosu+2cos3u3cos5u5]0π
I=127[2+223+25]0π
Hence
I=12315=1120

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