Integrating cos(px) from 1 to 2 with a constant p

[winston2020]

Homework Statement

Solve the following Integral:
\int_{1}^2cos(px)dx
where p is a constant

Homework Equations

The Attempt at a Solution

I'm totally lost here...

 

[PowerIso]

This isn't to bad. So, let u = px. du = pdx. So can you take it from there?

 

[winston2020]

This isn't to bad. So, let u = px. du = pdx. So can you take it from there?

Not really... what is du = pdx? du is the same as \frac{d}{dx}u right? But why is that useful? And what is pdx?

 

[BlackJack]

It's just a subsitution.

If du = p*dx then dx = du / p. Now integrate normally and at the end re-substitute.

 

[snipez90]

Substitution is important and knowing how to u-sub is the key to many integrals. But sometimes knowing that integration and differentiation are inverse operations allows you to guess the antiderivative.

What is the antiderivative of cos(x)? Where should the p be included? How do constants work when differentiating/integrating? You'll see that these questions aren't very hard to answer and it's more about thinking than just a routine substitution (though u-sub can get pretty tricky sometimes).

 

[winston2020]

Ok, should it go something like this?:

\int_{1}^2cos(px)dx

Let u = px

Therefore, du = pdx

And, dx = \frac{du}{p}

So,

\int_{1}^2cos(px)dx = \int_{1}^2cos(u)\frac{du}{p}
= \frac{sin(u)}{p} + c

Is that correct?

 

[winston2020]

anyone?

 

[physicsnoob93]

It would be more convenient to pull the 1/p out of the integral. Your solution seems correct.

 

[Mute]

You've evaluated the indefinite integral, but you still need to evaluate it at the limits you're given before the problem is complete.

i.e.,

\int_a^b f(x)~dx = F(b) - F(a)

where F(x) is the antiderivative of f(x).

 

[olgranpappy]

So,

\int_{1}^2cos(px)dx = \int_{1}^2cos(u)\frac{du}{p}
= \frac{sin(u)}{p} + c

Is that correct?

Almost, but

\int_{1}^2cos(px)dx \neq \int_{1}^2cos(u)\frac{du}{p}

...the limits are wrong on the RHS.

...in addition, you can easily see that your final equal sign is wrong by considering what happens if p=0. The answer should be 1... but your wrong answer is infinite at that value of p.

 

[HallsofIvy]

It is true that \int cos(px)dx= (1/p)sin(px)+ C. In order to do the definite integral, evaluating at the limits of integration, you can either write
\int_1^2 cos(px)dx= \left \frac{1}{p}sin(px)+ C\left|_1^2
and evaluate at x= 1 and 2 or you can make the substitution u= px so that when x= 1, u= p and when x= 2, u= 2p and write
\int_1^2 cos(px)dx= \frac{1}{p}\int_p^{2p} cos(u)du=\left \frac{1}{p}sin(u)\right|_p^{2p}

In more complicated problems you might have to make several substitutions and then it is better to change the limits of integration as you go (second method).