Can You Derive the Formula for cos^(n-1)(x) Using a Recursive Method?

[Patjamet]

I'm working on a Maths assignment and one of the problem solving questions is to prove:
\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

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 cos^{n-1}(x)[/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.) :)

y=cos^{n-1}(x)Allow cos(x) to equal u\frac{dy}{dx}=u^{n-1}....1


\frac{du}{dx}=-sin(x)

\frac{du}{-sin(x)}=dx.....2Sub 2 into 1\frac{dy}{\frac{du}{-sin(x)}}=u^{n-1}

\frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x)

\frac{dy}{du}=(n-1)u^{n-2}.-sin(x)

But u = cosx

\frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x)

 

[HallsofIvy]

I'm working on a Maths assignment and one of the problem solving questions is to prove:
\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

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 cos^{n-1}(x)[/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.) :)

y=cos^{n-1}(x)


Allow cos(x) to equal u


\frac{dy}{dx}=u^{n-1}....1

No, you haven't done the derivative yet. y= u^{n-1}
and then dy/dx= (n-1)u^{n-2} <br /> <br /> [/quote]\frac{du}{dx}=-sin(x)<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> \frac{du}{-sin(x)}=dx.....2 </div> </div> </blockquote> Why are you doing this? Use the chain rule: (du/dx)= (du/dy)(dy/dx)<br /> That gives you )n-1)u^{n-2}(-sin(x))= -(n-2)sin(x)cos^{n-2}(x)&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; Sub 2 into 1&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; \frac{dy}{\frac{du}{-sin(x)}}=u^{n-1} &lt;br /&gt; &lt;br /&gt; \frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x) &lt;br /&gt; &lt;br /&gt; \frac{dy}{du}=(n-1)u^{n-2}.-sin(x) &lt;br /&gt; &lt;br /&gt; But u = cosx&lt;br /&gt; &lt;br /&gt; \frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x) &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&lt;br /&gt; But I see no reason to ask about the &amp;quot;n-1&amp;quot; power. For your problem you want to look at cos^n(x). Again, letting u= cos(x), y= u^n so dy/du= n u^{n-1} and du/dx= - sin(x). So dy/dx= (dy/du)(du/dx)= n u^{n-1}(-sin(x))= -n sin(x)cos^{n-1}(x).&lt;br /&gt; &lt;br /&gt; With a little practice, you should be able to do that with actually writing down the &amp;quot;u&amp;quot; substitution: to differentiate cos^n(x) think &amp;quot;The &amp;quot;outer function&amp;quot; is a power so the derivative is n cos^{n-1}(x) and then I multiply by &lt;b&gt;its&lt;/b&gt; derivative, -sin(x)&amp;quot;.

 

[Patjamet]

Doh, the chain rule.

Thank you very much, looks like I have a lot of revision to do!