How do you find the antiderivative of $\int cos (π t) cos (sin (π t)) d t$ $int cos(pit)cos(sin(pit))dt$ ?

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How do you find the antiderivative of $\int cos (π t) cos (sin (π t)) d t$ $int cos(pit)cos(sin(pit))dt$ ?

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Answer 1 · 2017-01-14T04:04:33

$\frac{1}{π} sin (sin (π t)) + C$ $1/pisin(sin(pit))+C$

Explanation:

Let's try to simplify the trig function within the trig function by letting $u$ $u$ be the inside function, that is, $u = sin (π t)$ $u=sin(pit)$ . Differentiating this shows that $d u = π cos (π t) d t$ $du=picos(pit)dt$ . (Recall to use the chain rule.)

We currently have $cos (π t) d t$ $cos(pit)dt$ in the integral, so we need the factor of $π$ $pi$ .

$\int cos (π t) cos (sin (π t)) d t = \frac{1}{π} \int cos (sin (π t)) \cdot π cos (π t) d t$ $intcos(pit)cos(sin(pit))dt=1/piintcos(sin(pit))*picos(pit)dt$

Substituting in:

$= \frac{1}{π} \int cos (u) d u$ $=1/piintcos(u)du$

This is a common integral:

$= \frac{1}{π} sin (u) + C$ $=1/pisin(u)+C$

Substituting back in $u = sin (π t)$ $u=sin(pit)$ :

$= \frac{1}{π} sin (sin (π t)) + C$ $=1/pisin(sin(pit))+C$

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How do you find the antiderivative of $\int cos (π t) cos (sin (π t)) d t$ $int cos(pit)cos(sin(pit))dt$ ?

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Explanation:

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How do you find the antiderivative of $\int cos (π t) cos (sin (π t)) d t$ $int cos(pit)cos(sin(pit))dt$ ?