Question

Can you apply double angle to this integral ? - yes its a stupid question .

Answer 1

If you have \(\cos(2\theta)\) or \(\sin(2\theta)\) you can use double angle.
However, you really only need substitution.

Answer 2

\[Average Power Density =\frac{ 1 }{ T }*\int\limits_{0}^{T}A* e ^{-\alpha*z} \cos^2(\omega*t -\beta*z)dt \]

Answer 3

You use half-angle formula for these ones.

Answer 4

I originally was going with the forms of integral in the back of the book . But it became more cumbersome to solve .

Answer 5

Thanks Wio

Answer 6

It looks like you have a lot of constants with respect to \(t\).

Answer 7

I would start with \(u=\omega t-\beta z\). Linear substitutions are cheap and easy.
You'll end up with an integral of the from\[
C\int\cos^2(u)~du
\]where \(C\) is a constant with respect to \(t\) and \(u\).

Answer 8

= (t + 1/u sin (u)) should should be the result of the integral ?

Answer 9

times the constant' of course

Answer 10

The result won't have two varibles

Answer 11

\[
\cos^2(u) = \frac{1+\cos(2u)}{2}
\]

Answer 12

right , .. C/2 [ t + 1/(2u)* sin(2u) ] , I know you said that there should not be two variables ? This is what I see when evaluating

Answer 13

How did you get \(t\)? There is only \(u\)

Answer 14

C/2 [ u + 1/(2u)* sin(2u) ] ? sorry I accidently put a t

Answer 15

\[
\int\frac{\cos(2u)}{2}du = \int \frac{\cos(v)}{4}dv = \Bigg| \frac{\sin(v)}{4} = \frac 14 \Bigg|\sin(2u)
\]

Answer 16

Thank you . working on it ...

Answer 17

I think I got the wrong solution . Z remains a constant .