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

[karush]

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

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

From there?

 

[topsquark]

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

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

 

[karush]

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

 

[MarkFL]

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 $x$:

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

 

[karush]

Thanks again
I'm slowly seeing the magic in this