Is:
$\displaystyle COS^2(5u)$
The same as:
$\displaystyle COS(5u) * COS(5u)$
I am having real problems with double angle formulas and integration.
Hope someone can answer this for me?
You mention a double angle identity, so does that mean you wanted to find cos (10u)?
$\displaystyle \begin{aligned}
cos\,10u&=cos(2\cdot 5u)\\
&=cos^2\, 5u - sin^2\, 5u
\end{aligned}$
That's where $\displaystyle cos^2(5u)$ would come in I guess; and yes, it does equal $\displaystyle cos\,5u \cdot cos\,5u$.
Hello MaverickUK82Yes, it is!
If you want to integrate $\displaystyle \cos^25u$, use the formula:I am having real problems with double angle formulas and integration.
Hope someone can answer this for me?
$\displaystyle \cos2x = 2\cos^2x-1$Re-arranged to give:
$\displaystyle \cos^2x = \frac12(\cos2x+1)$So:
$\displaystyle \cos^25u = \frac12(\cos10u+1)$which is then very straightforward to integrate.
Grandad
If you have $\displaystyle \frac{1}{2}\cos{14x}+\frac{1}{2}$, then after integrating you get $\displaystyle \frac{1}{28}\sin{14x}+\frac{1}{2}x+k$. The fact used here is that the integral of $\displaystyle {\alpha}\cos{\beta{x}}$ is $\displaystyle \frac{\alpha}{\beta}\sin{\beta{x}}+k$, which might as well be what you were missing.
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} = $\displaystyle \frac{x}{y}$ or \dfrac{x}{y} = $\displaystyle \dfrac{x}{y}$
My function to integrate (don't know how to do integral symbol on here)
$\displaystyle cos^2(7x) dx$
Use this double angle formula:
$\displaystyle cos(2a)=2cos^2(a)-1$
Rearranged:
$\displaystyle \frac{1}{2}(cos(2a)+1) = cos^2(a))$
Substitute values:
$\displaystyle \frac{1}{2}(cos(2(7x))+1)$
Evaluate:
$\displaystyle \frac{1}{2}(cos(14x)+1)$
Integrate:
$\displaystyle \frac{1}{2}(\frac{1}{14}sin(14x)+x)$
Simplify:
$\displaystyle \frac{1}{28}(sin(14x)+x+c)$
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.
What about this:
$\displaystyle \frac{sin(14x)+x}{28}+c$
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)