Calculus Examples

$\int_{- 1}^{1} - 1 \cos (n π t) d t$

Step 1

Rewrite $- 1 \cos (n π t)$ as $- \cos (n π t)$ .

$\int_{- 1}^{1} - \cos (n π t) d t$

Step 2

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$- \int_{- 1}^{1} \cos (n π t) d t$

Step 3

Let $u = n π t$ . Then $d u = π n d t$ , so $\frac{1}{π n} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = n π t$ . Find $\frac{d u}{d t}$ .

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Differentiate $n π t$ .

$\frac{d}{d t} [n π t]$

Since $n π$ is constant with respect to $t$ , the derivative of $n π t$ with respect to $t$ is $n π \frac{d}{d t} [t]$ .

$n π \frac{d}{d t} [t]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$n π \cdot 1$

Simplify the expression.

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Multiply $n$ by $1$ .

$n π$

Reorder the factors of $n π$ .

$π n$

Substitute the lower limit in for $t$ in $u = n π t$ .

$u_{lower} = n π \cdot - 1$

Simplify.

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Move $- 1$ to the left of $n π$ .

$u_{lower} = - 1 \cdot (n π)$

Rewrite $- 1 (n π)$ as $- (n π)$ .

$u_{lower} = - n π$

Substitute the upper limit in for $t$ in $u = n π t$ .

$u_{upper} = n π \cdot 1$

Multiply $n$ by $1$ .

$u_{upper} = n π$

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - n π u_{upper} = n π$

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$- \int_{- n π}^{n π} \cos (u) \frac{1}{π n} d u$

Step 4

Combine $\cos (u)$ and $\frac{1}{π n}$ .

$- \int_{- n π}^{n π} \frac{\cos (u)}{π n} d u$

Step 5

Since $\frac{1}{π n}$ is constant with respect to $u$ , move $\frac{1}{π n}$ out of the integral.

$- (\frac{1}{π n} \int_{- n π}^{n π} \cos (u) d u)$

Step 6

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$- \frac{1}{π n} \sin (u)]_{- n π}^{n π}$

Step 7

Combine $\sin (u)]_{- n π}^{n π}$ and $\frac{1}{π n}$ .

$- \frac{\sin (u)]_{- n π}^{n π}}{π n}$

Step 8

Evaluate $\sin (u)$ at $n π$ and at $- n π$ .

$- \frac{\sin (n π) - \sin (- n π)}{π n}$