Solving Physics Problem: Integral from 0 to pi (sin(theta)cos(theta))

[pt176900]

I am attempting to solve a physics problem, and in the process am getting tripped up by an integral:

integrate from 0 to pi (sin(theta)cos(theta) d_theta)

sorry, I don't know how to use the symbols =/

at any rate, I substitute U for sin theta, du is cos theta, and integrate and of course I get sin^2(theta)/2 which of course equals 1 - cos 2(theta), evaluated from zero to pi, which gives me an answer of zero.

In the context (potential on the surface of a sphere), this doesn't make any sense - so obviously my math must be wrong somewhere. if someone can point it out, then i'd be much obliged.

Thanks

 

[quantumdude]

You didn't make any mistake evaluating the integral, so if you think your answer is wrong, then it's got to be in the integral itself. Can you state the full problem here?

 

[M98Ranger]

for sharing your problem with us! Integrals can definitely be tricky, but with a little bit of practice and understanding, you'll be able to solve them with ease.

First, let's take a look at the integral you're trying to solve: ∫sin(θ)cos(θ)dθ from 0 to π. Your substitution of u = sin(θ) and du = cos(θ)dθ is a great start, but there is a small mistake in your integration. When substituting, you also need to change the limits of integration. In this case, the limits would become u = sin(0) = 0 and u = sin(π) = 0. This means that your new integral would be ∫u du from 0 to 0, which is equal to 0.

To solve this integral correctly, we can use the trigonometric identity sin(2θ) = 2sin(θ)cos(θ). This allows us to rewrite the integral as ∫sin(θ)cos(θ)dθ = ∫(1/2)sin(2θ)dθ. Now, using the substitution u = 2θ and du = 2dθ, we can rewrite the integral once again as ∫(1/4)sin(u)du. This integral is much simpler to solve, and the final answer is (1/4)(-cos(u)) evaluated from 0 to 2π, which gives us a final answer of (1/4)(-cos(2π) + cos(0)) = 0.

As you can see, the mistake in your integration led to an incorrect answer. By properly substituting and changing the limits of integration, we were able to solve the integral and get the correct answer of 0. Keep practicing and don't be afraid to ask for help when needed. Good luck with your physics problem!