JH K.

asked • 08/13/20# I have a calculus question

I came across this integral while doing a physics question.

Is it possible to calculate:

∫cos(θ(t)) dt

with θ(t) as function in terms of t.

I'm hoping to somehow remove the integral so it is easier to work with. θ(t) is continuous and differentiable everywhere, if it helps.

Note:

I tried using the Maclaurin expansion of cos(x), but I got stuck at finding:

∫ θ(t)^{2k+1 }dt

for k ∈ [0,∞]

## 2 Answers By Expert Tutors

For computation (if you have upper and lower limits of integration), I imagine that because your integrand is periodic, one of the most accurate and simple ways to integrate would be __trapezoidal method__.

If you want an analytical approximation, it seems like what you are trying with the Maclaurin series is reasonable. If Theta is hard to work with, definitely use something like Mathematica to make it easier.

A third option would be to explore __asymptotic series methods.__

Hope this helps a bit!

Jeffrey K. answered • 08/13/20

Together, we build an iron base in mathematics and physics

Hi JHK:

Yes, it is usually possible to solve such an integral. It's simply the integral of a function of a function.

So, we substitute u = θ(t), calculate du = dθ/dt . dt, rearrange to get dt in terms of du, and substitute back into the original integral, solve the new (simpler) integral and then substitute back for u in teems of t.

Example: suppose the integral is: ∫cos(3t + 6) dt.

Then, u = θ(t) = 3t + 6^{ }=> dθ/dt = 3 and du = 3 dt => dt = (1/3) du

Substitute into the original integral: ∫cos(3t + 6) dt = (1/3) ∫cos u du

= -(1/3) sin u + C for arbitrary constant C

= -(1/3) sin (3t + 6) + C

Easy! No need to expand with Maclaurin (unless the new integral can't be done in terms of simple algebraic functions.)

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Douglas B.

08/13/20