Integrating $\cos(u)\cos(\sin(u))$

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Discussion Overview

The discussion revolves around the integration of the function $$\cos\left({\pi t}\right)\cos\left({\sin\left({\pi t}\right)}\right)$$ with respect to $$t$$. Participants explore various substitution methods and approaches to simplify the integral.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant suggests the substitution $$u=\pi t$$ and $$du=\pi \ dt$$ to simplify the integral.
  • Another participant proposes an additional substitution of $$x = \pi t$$ followed by $$u = \sin(x)$$ and $$du = \cos(x) dx$$.
  • There is a claim that the integral simplifies to $$\int\cos\left({u}\right)du$$, leading to a result involving the sine function.
  • One participant reports that their calculator returned a specific result for the integral, which includes back-substitutions for $$u$$ and $$x$$.

Areas of Agreement / Disagreement

Participants appear to agree on the substitution methods and the resulting integral form, but there is no consensus on the final answer as it is derived from different approaches and interpretations.

Contextual Notes

Some steps in the integration process are not fully detailed, and there may be assumptions regarding the validity of the substitutions that are not explicitly stated.

karush
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$$\int\cos\left({\pi t}\right)\cos\left({\sin\left({\pi t}\right)}\right) dt$$

Couldn't find a way to simplify this so
$$u=\pi t$$
$$du=\pi \ dt$$

From there?
 
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karush said:
$$\int\cos\left({\pi t}\right)\cos\left({\sin\left({\pi t}\right)}\right) dt$$

Couldn't find a way to simplify this so
$$u=\pi t$$
$$du=\pi \ dt$$

From there?
First let x = pi t. Then try u = sin(x), du = cos(x) dx

-Dan
 
So does this go to
$$\int\cos\left({u}\right)du $$

My TI nspire returned
$$\frac{\sin\left({\sin\left({\pi t}\right)}\right)}{\pi}+C$$

For the final answer
 
karush said:
So does this go to
$$\int\cos\left({u}\right)du $$

My TI nspire returned
$$\frac{\sin\left({\sin\left({\pi t}\right)}\right)}{\pi}+C$$

For the final answer

Well...what you have after Dan's suggested subs is

$$I=\frac{1}{\pi}\int \cos(u)\,du=\frac{1}{\pi}\sin(u)+C$$

Back-substitute for $u$:

$$I=\frac{1}{\pi}\sin\left(\sin(x)\right)+C$$

Back-substitute for $x$:

$$I=\frac{1}{\pi}\sin\left(\sin(\pi t)\right)+C$$
 
Thanks again
I'm slowly seeing the magic in this
 

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