How do you find the integral of $sin (2 π t) d t$ $sin(2 pi t) dt$ ?

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How do you find the integral of $sin (2 π t) d t$ $sin(2 pi t) dt$ ?

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Answer 1 · 2016-03-17T03:08:01

$\frac{- cos (2 π t)}{2 π} + C$ $(-cos(2pit))/(2pi)+C$

Explanation:

We have:

$\int sin (2 π t) d t$ $intsin(2pit)dt$

Substitute:

$u = 2 π t \Rightarrow d u = 2 π d t$ $u=2pit" "=>" "du=2pidt$

In order to have $d u$ $du$ in our integral expression, we must multiply the inside by $2 π$ $2pi$ . However, we also must balance this by multiplying the outside by $1 / 2 π$ $1"/"2pi$ .

$= \frac{1}{2 π} \int sin (2 π t) \cdot 2 π d t$ $=1/(2pi)intsin(2pit)*2pidt$

Substituting, we obtain:

$= \frac{1}{2 π} \int sin (u) d u$ $=1/(2pi)intsin(u)du$

This is a common integral:

$= - \frac{1}{2 π} cos (u) + C$ $=-1/(2pi)cos(u)+C$

Back substituting for $u$ $u$ :

$= \frac{- cos (2 π t)}{2 π} + C$ $=(-cos(2pit))/(2pi)+C$

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How do you find the integral of $sin (2 π t) d t$ $sin(2 pi t) dt$ ?

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