Integration of a complicated sort

[mateomy]

Homework Statement

Here's the original problem in all of its glory...

$$\int_0^{T/2} \,\sin((2\pi(t)/T) - \alpha) \, dt$$

Homework Equations

U Substitution
Trig identities (possibly)

The Attempt at a Solution

Using U substitution I've got u=$2\pi(t)/T -\alpha$

Which means that du= $2\pi/T dt$

Which also means that dt= $Tdu/2\pi$

and resetting the limits of integration I have an upper bound at $2\pi - \alpha$ and a lower bound of $-\alpha$.

So now my little equation looks like this...

$$\int_{-\alpha}^{2\pi - \alpha} \, \sinu \,\frac{T}{2\pi}\,du$$

pulling out the $T/2\pi$...

$$\frac{T}{2\pi}\int_{-\alpha}^{2\pi -\alpha} \,\sinu \,du$$

which using the Fundamental Theorem of Calculus I can turn to...

$$\frac{T}{2\pi} \, (-\cos(u))$$

evaluating at the upper minus the lower bound (previously stated)

Which looks something like this...

$$-\frac{T}{2\pi}\Big(\cos[\,2\pi\,-\,\alpha] - \cos[-\alpha]\Big)$$

Can I use a trigonometric identity to break this thing apart?
When I try to do that I get this...

$$-\frac{T}{2\pi}\,\Big(\cos(2\pi)\cos(\alpha)\,+\,\sin(2\pi)sin(\alpha)\,+\,\cos(\alpha)\Big)$$

All in all I end up with a final answer of

$$-\frac{T}{2\pi}\,2\cos(\alpha)$$

I have no answer to check my work, so that's why I am posting it here for everyone to point out my probable mistakes and give me some advice. Thanks a ton for the help/suggestions.

 

[mateomy]

Sorry if my Latex is screwed up in some places.And on second glance...would this end up being a 0 in the end because the Integration with the new U substitution turns out to be an odd function? Or am I closer to being right the first time?

 

[Mark44]

Fixed your first integral...I think.

Homework Statement

Here's the original problem in all of its glory...

$$\int_0^{T/2} sin(2\pi(t)/T - \alpha) \, dt$$

Homework Equations

U Substitution
Trig identities (possibly)

The Attempt at a Solution

Using U substitution I've got u=$2\pi(t)/T -\alpha$

Which means that du= $2\pi/T dt$

Which also means that dt= $Tdu/2\pi$

and resetting the limits of integration I have an upper bound at $2\pi - \alpha$ and a lower bound of $-\alpha$.

So now my little equation looks like this...

$$\int_-\alpha^(2\pi - \alpha) \, \sinu \,\frac{T}{2\pi}\,du$$

pulling out the $T/2\pi$...

$$\frac{T}{2\pi}\int_-\alpha^(2\pi -\alpha) \,\sinu \,du$$

which using the Fundamental Theorem of Calculus I can turn to...

$$\frac{T}{2\pi} \, -\cos(u)$$

evaluating at the upper minus the lower bound (previously stated)

Which looks something like this...

$$-\frac{T}{2\pi}\Big(\cos[\,2\pi\,-\,\alpha] - \cos[-\alpha]\Big)$$

Can I use a trigonometric identity to break this thing apart?
When I try to do that I get this...

$$-\frac{T}{2\pi}\,\Big(\cos(2\pi)\cos(\alpha)\,+\,\sin(2\pi)sin(\alpha)\,+\,\cos(\alpha)\Big)$$

All in all I end up with a final answer of

$$-\frac{T}{2\pi}\,2\cos(\alpha)$$

I have no answer to check my work, so that's why I am posting it here for everyone to point out my probable mistakes and give me some advice. Thanks a ton for the help/suggestions.

 

[mateomy]

Yeah you fixed it. I'll try to work out the other integrals using your code as an example.

THANKS!

 

[Mark44]

Homework Statement

Here's the original problem in all of its glory...

$$\int_0^{T/2} \,\sin((2\pi(t)/T) - \alpha) \, dt$$

Homework Equations

U Substitution
Trig identities (possibly)

The Attempt at a Solution

Using U substitution I've got u=$2\pi(t)/T -\alpha$

Which means that du= $2\pi/T dt$

Which also means that dt= $Tdu/2\pi$

and resetting the limits of integration I have an upper bound at $2\pi - \alpha$ and a lower bound of $-\alpha$.

So now my little equation looks like this...

$$\int_{-\alpha}^{2\pi - \alpha} \, \sinu \,\frac{T}{2\pi}\,du$$

pulling out the $T/2\pi$...

$$\frac{T}{2\pi}\int_{-\alpha}^{2\pi -\alpha} \,\sinu \,du$$

which using the Fundamental Theorem of Calculus I can turn to...

$$\frac{T}{2\pi} \, (-\cos(u))$$

evaluating at the upper minus the lower bound (previously stated)

Which looks something like this...

$$-\frac{T}{2\pi}\Big(\cos[\,2\pi\,-\,\alpha] - \cos[-\alpha]\Big)$$

You have an error in the line above. Inside the parentheses it should be cos(pi - alpha) - cos(-alpha)

cos(pi - alpha) = -cos(alpha) and cos(-alpha) = cos(alpha)

Can I use a trigonometric identity to break this thing apart?
When I try to do that I get this...

$$-\frac{T}{2\pi}\,\Big(\cos(2\pi)\cos(\alpha)\,+\,\sin(2\pi)sin(\alpha)\,+\,\cos(\alpha)\Big)$$

All in all I end up with a final answer of

$$-\frac{T}{2\pi}\,2\cos(\alpha)$$

I have no answer to check my work, so that's why I am posting it here for everyone to point out my probable mistakes and give me some advice. Thanks a ton for the help/suggestions.

I end up with (T/pi)cos(alpha).

 

[mateomy]

Okay, so I am close...

Im looking at the point where I pulled the antiderivative of sin(u), which I thought was a (neg)cos(u). So wouldn't I want to pull it all the way out in front of my constant $\frac{T}{2\pi}$?

Im also lost on how you reached a pi as opposed to the 2pi I have come to?

But seriously, thanks.

 

[mateomy]

Im also lost on how you reached a pi as opposed to the 2pi I have come to?

But seriously, thanks.

I meant that in sense of what was in the parenthesis, not the final answer; because, correct me if I am wrong but the 2pi in the denominator and 2cos would simplify to a pi in the denominator and a singular cosine.

 

[Mark44]

When you evaluate cos(2pit/T - alpha) at t = T/2, you get cos((2pi/T)(T/2) - alpha). The 2's cancel and the T's cancel, so you're left with cos(pi - alpha).

 

[mateomy]

Hmmmmmm