# Thread: Quick one regardin Cos^2... (Needed for integration)

1. ## Quick one regardin Cos^2... (Needed for integration)

Is:
$COS^2(5u)$

The same as:
$COS(5u) * COS(5u)$

I am having real problems with double angle formulas and integration.

Hope someone can answer this for me?

2. You mention a double angle identity, so does that mean you wanted to find cos (10u)?
\begin{aligned}
cos\,10u&=cos(2\cdot 5u)\\
&=cos^2\, 5u - sin^2\, 5u
\end{aligned}

That's where $cos^2(5u)$ would come in I guess; and yes, it does equal $cos\,5u \cdot cos\,5u$.

3. Hello MaverickUK82
Originally Posted by MaverickUK82
Is:
$COS^2(5u)$

The same as:
$COS(5u) * COS(5u)$
Yes, it is!

I am having real problems with double angle formulas and integration.

Hope someone can answer this for me?
If you want to integrate $\cos^25u$, use the formula:
$\cos2x = 2\cos^2x-1$
Re-arranged to give:
$\cos^2x = \frac12(\cos2x+1)$
So:
$\cos^25u = \frac12(\cos10u+1)$
which is then very straightforward to integrate.

4. Brilliant. I was half way there then!

Thanks guys!

5. or maybe not...

I've tried integrating this using the above method:

$
f(x)=cos^2(7u)
$

and got this:

$1/2 sin(14x)+x+c$

6. If you have $\frac{1}{2}\cos{14x}+\frac{1}{2}$, then after integrating you get $\frac{1}{28}\sin{14x}+\frac{1}{2}x+k$. The fact used here is that the integral of ${\alpha}\cos{\beta{x}}$ is $\frac{\alpha}{\beta}\sin{\beta{x}}+k$, which might as well be what you were missing.

7. I wrote that completely wrong:

I started with this:

$f(x)=cos^2(7x)$

and integrated to get this:

$1/2 sin(14x)+ 1/2 x + c$

sorry, don't know how to get the 1/2 as fractions on here.

is that right?

8. Not quite but nearly, you are just missing some constant in the integral. Can you show the steps you follow in order to get to this result ?

Also, to write fractions in LaTeX you use the command \frac{x}{y} = $\frac{x}{y}$ or \dfrac{x}{y} = $\dfrac{x}{y}$

9. I can't remember how I got there now, exactly and my working out is at home (I'm at work).
I'll have another go at it during my lunch break and post my working then!

10. My function to integrate (don't know how to do integral symbol on here)
$cos^2(7x) dx$

Use this double angle formula:
$cos(2a)=2cos^2(a)-1$

Rearranged:
$\frac{1}{2}(cos(2a)+1) = cos^2(a))$

Substitute values:
$\frac{1}{2}(cos(2(7x))+1)$

Evaluate:
$\frac{1}{2}(cos(14x)+1)$

Integrate:
$\frac{1}{2}(\frac{1}{14}sin(14x)+x)$

Simplify:
$\frac{1}{28}(sin(14x)+x+c)$

11. Or written like this:

$\frac{sin(14x)+x}{28}+c$

12. Originally Posted by MaverickUK82
My function to integrate (don't know how to do integral symbol on here)
$cos^2(7x) dx$

Use this double angle formula:
$cos(2a)=2cos^2(a)-1$

Rearranged:
$\frac{1}{2}cos(2a)+1 = cos^2(a)$ Mr F says: Wrong. It's $\frac{1}{2} [ cos(2a)+1] = cos^2(a)$

Substitute values:
$\frac{1}{2}cos(2(7x))+1$ Mr F says: Wrong. It's $\frac{1}{2} [cos(2(7x))+1]$

Evaluate:
$\frac{1}{2}cos(14x)+1$ Mr F says: Wrong. It's $\frac{1}{2} [cos(14x)+1]$

Integrate:
$\frac{1}{2}(\frac{1}{14}sin(14x)+x)$ Mr F says: Correct. I see the brackets that were missing in all of the above finally make an appearance.

Simplify:
$\frac{1}{28}(sin(14x)+x+c)$
Your final answer is still wrong. Does it expand out to give the previous correct line of working? It would save everyone's time (including your own) if you took greater care with basic algebra.

And take greater care with brackets.

13. Mr. F,
Sorry about the brackets, it's my poor use of LaTeX that's the problem as opposed to me missing them. I typed it out in a hurry because I'm at work. I didn't realise they were missing. Sorry to waste your time.

$\frac{sin(14x)+x}{28}+c$

14. Originally Posted by MaverickUK82
Mr. F,
Sorry about the brackets, it's my poor use of LaTeX that's the problem as opposed to me missing them. I typed it out in a hurry because I'm at work. I didn't realise they were missing. Sorry to waste your time.

$\frac{sin(14x)+x}{28}+c$
You had the correct answer in the second last line of that other post. Nothing else you have written after it is correct. So that I am not repeating myself, I will say take much greater care with your basic algebra.

(I just do not understand how you can possibly propose this or the other final answer as being equivalent to that correct second last line)

15. I understand now. I can't multiply the 1/2 by the 1/14. Sorry.

the +C bit at the end is right though?

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