By considering, \(\displaystyle\int{{\sin}^{{n}}{\left({x}\right)}}{{\cos}^{{3}}{\left({x}\right)}}{\left.{d}{x}\right.}\) Prove that: \(\displaystyle{\frac{{{{\sin}^{{8}}{\left({x}\right)}}}}{{{8}}}}-{\frac{{{{\sin}^{{6}}{\left({x}\right)}}}}{{{6}}}}+{\frac{{{1}}}{{{24}}}}={\frac{{{{\cos}^{{8}}{\left({x}\right)}}}}{{{8}}}}-{\frac{{{{\cos}^{{6}}{\left({x}\right)}}}}{{{3}}}}+{\frac{{{{\cos}^{{4}}{\left({x}\right)}}}}{{{4}}}}\) I have integrated the

hinnikendvjg

hinnikendvjg

Answered question

2022-03-18

By considering,
sinn(x)cos3(x)dx
Prove that:
sin8(x)8sin6(x)6+124=cos8(x)8cos6(x)3+cos4(x)4
I have integrated the suggested integral successfully, that being:
sinn+1(x)n+1sinn+3(x)n+3+c
I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand sin2(x)=(1cos2(x)) but this seems far too tedious.

Answer & Explanation

Meadow Franco

Meadow Franco

Beginner2022-03-19Added 6 answers

Hint:
I=sin2m+1xcos2n+1x dx=(1cos2x)mcos2n+1xsinxdx
Set cosx=u
Again,
I=(1sin2x)nsin2m+1xcosxdx
Set sinx=v
Here 2n+1=3
Set 2m+2n+2=8,6,4  etc.
elementalfoxwqe

elementalfoxwqe

Beginner2022-03-20Added 3 answers

hint...You just need to consider
sin5xcos3xdx
either as
sin5x(1sin2x)cosxdx
or as
sinx(1cos2x)2cos3xdx
These will differ by a constant which you can evaluate using eg. x=0

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