Complex Function Formalism Please Help Integrals

[pLk]

The Integral is already solved although I can't seem to reach the same solution given. If I plug it into an online calculator I get the Correct answer but I dont understand how it reached this value when I attempt by hand I reach a different value. Can some one help guide me in the process it would be appreciated.
All values shown below are 100% correct


This is the Integral


The Value of T = 2*Pi/w



This is another way of representing the highlighted Yellow makes it easier to work on the equation.


And this is the Answer.




When I try to do it by hand on paper I receive the following it is correct but some where I'm doing something wrong that I'm not reaching the above answer am I stop short? skipping steps etc... help please

 

[Dr.Peterson]

The Integral is already solved although I can't seem to reach the same solution given. If I plug it into an online calculator I get the Correct answer but I dont understand how it reached this value when I attempt by hand I reach a different value. Can some one help guide me in the process it would be appreciated.
All values shown below are 100% correct


This is the Integral
View attachment 31590

The Value of T = 2*Pi/w



This is another way of representing the highlighted Yellow makes it easier to work on the equation.
View attachment 31592

And this is the Answer.

View attachment 31588


When I try to do it by hand on paper I receive the following it is correct but some where I'm doing something wrong that I'm not reaching the above answer am I stop short? skipping steps etc... help please

View attachment 31591

Please show us your work, so we can see where you might be going wrong. What do you mean when you say "it is correct but some where I'm doing something wrong"?

But maybe you've shown enough. What you show as your answer is an indefinite integral; what you have to do is a definite integral! Just finish up by replacing t with the limits of integration and simplifying.

 

[nasi112]

Instead of writing the solution with \(\displaystyle + \ C\), substitute the limits of integration \(\displaystyle \int_0^T\).

So, your answer should be

\(\displaystyle \frac{AB}{8\pi}[\sin(2wT + b + a) + 2\cos(b - a)wT - \sin(b + a)]\)

Now change \(\displaystyle T\) to \(\displaystyle \frac{2\pi}{w}\)

\(\displaystyle \frac{AB}{8\pi}[\sin(4\pi + b + a) + 4\pi \cos(b - a) - \sin(b + a)]\)

Use this identity,

\(\displaystyle \sin(4\pi + b + a) = \sin(4\pi)\cos(a + b) + \cos(4\pi)\sin(a + b) = 0 + \sin(a + b) = \sin(a + b)\)

Substitute this back into the answer, and you will get,

\(\displaystyle \frac{AB}{8\pi}[\sin(b + a) + 4\pi \cos(b - a) - \sin(b + a)] = \frac{AB}{8\pi}[4\pi \cos(b - a)] = \frac{AB}{2}\cos(b - a)\)

 

[pLk]

Nice Nasi! Thank I just saw angels sum rule and figured it out but you also explain it very beautifully Thank you!!!

 

[Dr.Peterson]

Rather than

\(\displaystyle \sin(4\pi + b + a) = \sin(4\pi)\cos(a + b) + \cos(4\pi)\sin(a + b) = 0 + \sin(a + b) = \sin(a + b)\)

you could just use the fact that the sine has period [imath]2\pi[/imath], that is, \(\displaystyle \sin(x + 2k\pi) = \sin(x)\), so \(\displaystyle \sin(4\pi + b + a) = \sin(b + a)\).

 

[nasi112]

Nice Nasi! Thank I just saw angels sum rule and figured it out but you also explain it very beautifully Thank you!!!

You're welcome

Rather than

you could just use the fact that the sine has period [imath]2\pi[/imath], that is, \(\displaystyle \sin(x + 2k\pi) = \sin(x)\), so \(\displaystyle \sin(4\pi + b + a) = \sin(b + a)\).

Indeed, professor, and this will be a big shortcut.