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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Richiie on May 31, 2010, 08:08:29 pm

Title: Anti-differentiation
Post by: Richiie on May 31, 2010, 08:08:29 pm: How would I find the integral of ..

$sin^{2}(x)cos^{4}(x)$

Thank you.
Title: Re: Anti-differentiation
Post by: Yitzi_K on May 31, 2010, 11:18:37 pm: ~~This appears to work...~~

Edit: it didn't work
Title: Re: Anti-differentiation
Post by: Richiie on May 31, 2010, 11:27:11 pm: So far I've done up to this point, and I'm just completely stumped.

$\frac {1}{8}\int_{}^{}sin^{2}(2x)(cos(2x)+1) dx$
Title: Re: Anti-differentiation
Post by: Yitzi_K on May 31, 2010, 11:31:04 pm: Yeh I also got to something like that, i couldn't see any way of going further.

~~Although I believe my post above to be the correct answer.~~
Title: Re: Anti-differentiation
Post by: m@tty on May 31, 2010, 11:35:34 pm: How does $sin^{2}(x)cos^{4}(x)=sin^2(x) \times -2sin(x)cos(x) \times cos^2(x)$ ?
Title: Re: Anti-differentiation
Post by: Richiie on May 31, 2010, 11:36:34 pm: I found these solutions, just don't know how they proceeded.

$\frac {1}{8}\int_{}^{}sin^{2}(2x)(cos(2x)+1) dx$

$\frac {1}{8}\int_{}^{}sin^{2}(2x)(cos(2x)) dx+ \frac {1}{8}\int_{}^{}sin^{2}(2x) dx$

$\frac {1}{16}\int_{}^{}sin^{2}(2x) d(sin(2x)) + \frac {1}{16}\int_{}^{}(1-cos(4x)) dx$ <--- This step has stumped me.

$\frac {1}{48}sin^{3}(2x) + \frac {x}{16} - \frac {1}{64}sin(4x) + C$
Title: Re: Anti-differentiation
Post by: Yitzi_K on May 31, 2010, 11:38:00 pm: Quote from: m@tty on May 31, 2010, 11:35:34 pm
How does $sin^{2}(x)cos^{4}(x)=sin^2(x) \times -2sin(x)cos(x) \times cos^2(x)$ ?

you're right i stuffed up... i'll blame tiredness as an excuse :-[ :-[
Title: Re: Anti-differentiation
Post by: Richiie on May 31, 2010, 11:43:18 pm: Quote from: Yitzi_K on May 31, 2010, 11:38:00 pm
Quote from: m@tty on May 31, 2010, 11:35:34 pm
How does $sin^{2}(x)cos^{4}(x)=sin^2(x) \times -2sin(x)cos(x) \times cos^2(x)$ ?

you're right i stuffed up... i'll blame tiredness as an excuse :-[ :-[

Your effort is much appreciated. =D
Title: Re: Anti-differentiation
Post by: m@tty on May 31, 2010, 11:46:55 pm: Quote from: Richiie on May 31, 2010, 11:36:34 pm
$\frac {1}{16}\int_{}^{}sin^{2}(2x) d(sin(2x)) + \frac {1}{16}\int_{}^{}(1-cos(4x)) dx$ <--- This step has stumped me.

It's a double angle formula.

$\cos(2x)=\cos^2(x)-\sin^2(x)$

And $\cos^2(x)+\sin^2(x)=1 \implies \cos^2(x)=1-\sin^2(x)$

So the above formula becomes $\cos(2x)=1-2\sin^2(x)$

So you have $\sin^2(2x)$

From the above double angle formula $\cos(4x)=1-2\sin^2(2x)$

$\implies \sin^2(2x)=\frac{1}{2}\left(1-\cos(4x)\right)$ .

Sub that in your integral and you get to the next step :D
Title: Re: Anti-differentiation
Post by: Richiie on May 31, 2010, 11:50:49 pm: Ohh, I think I've almost solved this equation.

Edit: Ahh, I've finally solved it. Thank you for everyone's contribution. :D
Title: Re: Anti-differentiation
Post by: m@tty on May 31, 2010, 11:52:41 pm: Quote from: Richiie on May 31, 2010, 11:50:49 pm
Ohh, I think I've almost solved this equation.

Isn't it already solved? Or are you trying to simplify it? ?
Title: Re: Anti-differentiation
Post by: Richiie on May 31, 2010, 11:55:00 pm: Quote from: m@tty on May 31, 2010, 11:52:41 pm
Quote from: Richiie on May 31, 2010, 11:50:49 pm
Ohh, I think I've almost solved this equation.

Isn't it already solved? Or are you trying to simplify it? ?

I just got those answers by googling. I just didn't know how to get it.
Title: Re: Anti-differentiation
Post by: m@tty on June 01, 2010, 12:00:45 am: Ahh, lol. Was confused for a second there. Are you still confused with the question?
Title: Re: Anti-differentiation
Post by: kakar0t on June 01, 2010, 12:34:25 am: could someone please do a full solution for confused people like myself? :)
Title: Re: Anti-differentiation
Post by: Mao on June 01, 2010, 02:56:18 am: $\sin^2(x)\cos^4(x) = \frac{1}{4}(2\sin (x)\cos(x))^2\cos^2(x) = \frac{1}{4}\sin^2(2x)\cos^2(x) = \frac{1}{4}\sin^2(2x)\left(\frac{1+\cos(2x)}{2}\right) = \frac{1}{8}\sin^2(2x)(1+\cos(2x))$

$I = \int \frac{1}{8}\sin^2(2x) (1+\cos(2x))\; dx = \frac{1}{8}\left(\int \sin^2(2x)\; dx + \int \sin^2 (2x)\cos(2x) \; dx\right)$

$\int \sin^2(2x)\; dx = \int \frac{1-\cos(4x)}{2} \; dx = \frac{1}{2} \left( x - \frac{1}{4}\sin(4x) \right) + C$

$\int \sin^2(2x) \cos(2x)\; dx = \int \frac{1}{2} u^2\; du = \frac{1}{6}\sin^3(2x) + C$ , where $u=\sin (2x)$

Thus, $I = \frac{1}{8} \left( \frac{1}{2} \left( x - \frac{1}{4}\sin(4x) \right) + \frac{1}{6}\sin^3(2x) \right) + C = \frac{x}{16} - \frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + C$
Title: Re: Anti-differentiation
Post by: /0 on June 01, 2010, 06:01:59 am: Quote from: Richiie on May 31, 2010, 11:36:34 pm
I found these solutions, just don't know how they proceeded.

$\frac {1}{8}\int_{}^{}sin^{2}(2x)(cos(2x)+1) dx$

$\frac {1}{8}\int_{}^{}sin^{2}(2x)(cos(2x)) dx+ \frac {1}{8}\int_{}^{}sin^{2}(2x) dx$

$\frac {1}{16}\int_{}^{}sin^{2}(2x) d(sin(2x)) + \frac {1}{16}\int_{}^{}(1-cos(4x)) dx$ <--- This step has stumped me.

$\frac {1}{48}sin^{3}(2x) + \frac {x}{16} - \frac {1}{64}sin(4x) + C$

That step is u-substitution. Some people just prefer not to write "u" and instead they keep the whole " $\sin{(2x)}$ ".
Title: Re: Anti-differentiation
Post by: Richiie on June 01, 2010, 08:03:07 pm: Quote from: m@tty on June 01, 2010, 12:00:45 am
Ahh, lol. Was confused for a second there. Are you still confused with the question?

Nope, finally solved it. xD