Difference in integral not understood

[Nikunj Parikh]

Hi

I am out of touch of maths for last 12 years and now reviving my math.
I am getting two different answers for the same integral.

I= Integral (sinx)^2(cosx)dx

method-1
If I proceed with (sinx)^2 = 1-(cosx)^2, I endup with answer sinx-((cosx)^4)/4)

method-2
with method of substitution u=sinx and hence du=cosxdx so the answer is ((sinx)^3)/3

Since both the methods look correct me, I am unable to find reason for this difference. Can some one help me

Attachment available from http://mathforum.org/kb/servlet/JiveServlet/download/125-2119055-7179463-617096/Graphical%20representation.doc

[Federico Zanolla]

For the first one you have:
integral cos(x) - cos^3(x).
Simply integrate term by term using this reduction formula on -cos^3(x)

integral cos^m(x) dx = [(sin(x) cos^(m-1)(x))/m] + [(m-1)/m] integral cos^(-2+m)(x) dx, where m = 3

Eventually you'll end up with the same result as per the second method.

[Nikunj Parikh]

Thanks Federico

Thanks for your time. Appreciate your help.

Regards
Nikunj

[Nikunj Parikh]

Thanks Louis

Appreciate your support and efforts.

Regards
Nikunj
- ----- Original Message -----
From: "Louis Talman" <tal...@mscd.edu>
To: "Nikunj Parikh" <nikunj...@gmail.com>
Cc: "Louis Talman" <tal...@mscd.edu>; <geometry...@support1.mathforum.org>
Sent: Monday, September 06, 2010 10:36 PM
Subject: Re: Difference in integral not understood

On Sep 6, 2010, at 6:11 AM, Nikunj Parikh wrote:

> If I proceed with (sinx)^2 = 1-(cosx)^2, I endup with answer
> sinx-((cosx)^4)/4)

Not if you do the resulting integral correctly. [(Cos[x])^4]/4 is *not* an
antiderivative for (Cos[x])^3.

- --Lou Talman
Department of Mathematical and Computer Sciences
Metropolitan State College of Denver

<http://rowdy.mscd.edu/%7Etalmanl>