Anti-differentiation

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Re: Anti-differentiation

« Reply #15 on: June 01, 2010, 06:01:59 am »

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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 " $\sin{(2x)}$ ".

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Richiie

Victorian
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Posts: 179
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School Grad Year: 2010

Re: Anti-differentiation

« Reply #16 on: June 01, 2010, 08:03:07 pm »

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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

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2009: Biology
2010: English, Mathematical Methods (CAS), Specialist Maths, Chemistry, Physics
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