# Some help with a relatively straight forward integral please!

• Jun 28th 2012, 05:43 AM
euphmorning
Some help with a relatively straight forward integral please!
Just trying to understand the answer in my book here...

given that cos(2x) = 2cos^2 x -1

then:

cos^2 x = 1/2(cos(2x) + 1). agreed.

my problem comes upon integrating cos^2 x..

i get thus far:

cos^2 x dx = 1/2(cos(2x) + 1).

=1/2(1/2 sin(2x) + x) + c

agreed. it is the next bit that is throwing me off..

distributing out the 1/2 i get:

1/4sin(2x) + 2x) + c

i'm sure i'm missing something terribly obvious here, but where does the the extra x (2x term) pop up from ?

• Jun 28th 2012, 05:49 AM
Prove It
Re: Some help with a relatively straight forward integral please!
Quote:

Originally Posted by euphmorning
Just trying to understand the answer in my book here...

given that cos(2x) = 2cos^2 x -1

then:

cos^2 x = 1/2(cos(2x) + 1). agreed.

my problem comes upon integrating cos^2 x..

i get thus far:

cos^2 x dx = 1/2(cos(2x) + 1).

=1/2(1/2 sin(2x) + x) + c

agreed. it is the next bit that is throwing me off..

distributing out the 1/2 i get:

1/4sin(2x) + 2x) + c

i'm sure i'm missing something terribly obvious here, but where does the the extra x (2x term) pop up from ?

The antiderivative of 1 is x. Also, when you should distribute you should get (1/2)x, not 2x.
• Jun 28th 2012, 05:54 AM
Reckoner
Re: Some help with a relatively straight forward integral please!
Quote:

Originally Posted by euphmorning
1/4sin(2x) + 2x) + c

i'm sure i'm missing something terribly obvious here, but where does the the extra x (2x term) pop up from ?

I think you may be missing a parenthesis there. I'm assuming that you're trying to factor out the $\displaystyle \textstyle\frac12?$

$\displaystyle \frac12\left(\frac12\sin2x + x\right) + C$

$\displaystyle =\frac12\left(\frac12\sin2x + \frac12\cdot2x\right) + C$

$\displaystyle =\frac12\cdot\frac12\left(\sin2x + 2x\right) + C$

$\displaystyle =\frac14\left(\sin2x + 2x\right) + C$
• Jun 28th 2012, 08:04 AM
euphmorning
Re: Some help with a relatively straight forward integral please!
[QUOTE=Reckoner;724839]I think you may be missing a parenthesis there. I'm assuming that you're trying to factor out the $\displaystyle \textstyle\frac12?$

$\displaystyle \frac12\left(\frac12\sin2x + x\right) + C$

$\displaystyle =\frac12\left(\frac12\sin2x + \frac12\cdot2x\right) + C$

$\displaystyle =\frac12\cdot\frac12\left(\sin2x + 2x\right) + C$

[tex]=\frac14\left(\sin2x + 2x\right) + C

Thanks alot! That sorted it out nicely :)