- #1
Patjamet
- 6
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I'm working on a Maths assignment and one of the problem solving questions is to prove:
[tex]\int cos^{8}(x)dx=\frac{1}{8}cos^{7}(x).sin(x)+\frac{7}{48}cos^{5}(x).sin(x)+\frac{35}{192}cos^{3}(x).sin(x)+\frac{35}{128}cos(x).sin(x)+\frac{35}{128}(x)+c[/tex]
I have looked through my textbook and found a good example which will help me, the book uses a thing they are calling a recursive formula. I've done some research and I've seen the same formula from lots of different information sources.
Exhibit A: http://calc101.com/deriving_reduction_2.html
[STRIKE]They have all given demonstrations on how to get the formula, but when I try to do it myself I run into a problem when differentiating the first function [tex]cos^{n-1}(x)[/tex][/STRIKE]
Nevermind I was integrating haha, but still; obviously I am rusty on my differentiation techniques so if someone could check this for me that would be fantastic (mainly concerned with my procedure for communication marks etc.) :)
[tex]y=cos^{n-1}(x)[/tex]Allow cos(x) to equal u[tex]\frac{dy}{dx}=u^{n-1}....1[/tex]
[tex]\frac{du}{dx}=-sin(x)[/tex]
[tex]\frac{du}{-sin(x)}=dx.....2[/tex]Sub 2 into 1[tex]\frac{dy}{\frac{du}{-sin(x)}}=u^{n-1}[/tex]
[tex]\frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x)[/tex]
[tex]\frac{dy}{du}=(n-1)u^{n-2}.-sin(x)[/tex]
But u = cosx
[tex]\frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x)[/tex]
[tex]\int cos^{8}(x)dx=\frac{1}{8}cos^{7}(x).sin(x)+\frac{7}{48}cos^{5}(x).sin(x)+\frac{35}{192}cos^{3}(x).sin(x)+\frac{35}{128}cos(x).sin(x)+\frac{35}{128}(x)+c[/tex]
I have looked through my textbook and found a good example which will help me, the book uses a thing they are calling a recursive formula. I've done some research and I've seen the same formula from lots of different information sources.
Exhibit A: http://calc101.com/deriving_reduction_2.html
[STRIKE]They have all given demonstrations on how to get the formula, but when I try to do it myself I run into a problem when differentiating the first function [tex]cos^{n-1}(x)[/tex][/STRIKE]
Nevermind I was integrating haha, but still; obviously I am rusty on my differentiation techniques so if someone could check this for me that would be fantastic (mainly concerned with my procedure for communication marks etc.) :)
[tex]y=cos^{n-1}(x)[/tex]Allow cos(x) to equal u[tex]\frac{dy}{dx}=u^{n-1}....1[/tex]
[tex]\frac{du}{dx}=-sin(x)[/tex]
[tex]\frac{du}{-sin(x)}=dx.....2[/tex]Sub 2 into 1[tex]\frac{dy}{\frac{du}{-sin(x)}}=u^{n-1}[/tex]
[tex]\frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x)[/tex]
[tex]\frac{dy}{du}=(n-1)u^{n-2}.-sin(x)[/tex]
But u = cosx
[tex]\frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x)[/tex]