Calculus Examples

$\frac{d v}{d t} = 8 t + \csc^{2} (t)$ , $v (\frac{π}{2}) = - 7$

Step 1

Rewrite the equation.

$d v = (8 t + \csc^{2} (t)) d t$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int d v = \int 8 t + \csc^{2} (t) d t$

Step 2.2

Apply the constant rule.

$v + C_{1} = \int 8 t + \csc^{2} (t) d t$

Step 2.3

Integrate the right side.

Tap for more steps...

Step 2.3.1

Split the single integral into multiple integrals.

$v + C_{1} = \int 8 t d t + \int \csc^{2} (t) d t$

Step 2.3.2

Since $8$ is constant with respect to $t$ , move $8$ out of the integral.

$v + C_{1} = 8 \int t d t + \int \csc^{2} (t) d t$

Step 2.3.3

By the Power Rule, the integral of $t$ with respect to $t$ is $\frac{1}{2} t^{2}$ .

$v + C_{1} = 8 (\frac{1}{2} t^{2} + C_{2}) + \int \csc^{2} (t) d t$

Step 2.3.4

Since the derivative of $- \cot (t)$ is $\csc^{2} (t)$ , the integral of $\csc^{2} (t)$ is $- \cot (t)$ .

$v + C_{1} = 8 (\frac{1}{2} t^{2} + C_{2}) - \cot (t) + C_{3}$

Step 2.3.5

Simplify.

Tap for more steps...

Step 2.3.5.1

Combine $\frac{1}{2}$ and $t^{2}$ .

$v + C_{1} = 8 (\frac{t^{2}}{2} + C_{2}) - \cot (t) + C_{3}$

Step 2.3.5.2

Simplify.

$v + C_{1} = 4 t^{2} - \cot (t) + C_{4}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$v = 4 t^{2} - \cot (t) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $\frac{π}{2}$ for $t$ and $- 7$ for $v$ in $v = 4 t^{2} - \cot (t) + K$ .

$- 7 = 4 {(\frac{π}{2})}^{2} - \cot (\frac{π}{2}) + K$

Step 4

Solve for $K$ .

Tap for more steps...

Step 4.1

Rewrite the equation as $4 {(\frac{π}{2})}^{2} - \cot (\frac{π}{2}) + K = - 7$ .

$4 {(\frac{π}{2})}^{2} - \cot (\frac{π}{2}) + K = - 7$

Step 4.2

Simplify the left side.

Tap for more steps...

Step 4.2.1

Simplify $4 {(\frac{π}{2})}^{2} - \cot (\frac{π}{2}) + K$ .

Tap for more steps...

Step 4.2.1.1

Simplify each term.

Tap for more steps...

Step 4.2.1.1.1

Apply the product rule to $\frac{π}{2}$ .

$4 \frac{π^{2}}{2^{2}} - \cot (\frac{π}{2}) + K = - 7$

Step 4.2.1.1.2

Raise $2$ to the power of $2$ .

$4 \frac{π^{2}}{4} - \cot (\frac{π}{2}) + K = - 7$

Step 4.2.1.1.3

Cancel the common factor of $4$ .

Tap for more steps...

Step 4.2.1.1.3.1

Cancel the common factor.

$4 \frac{π^{2}}{4} - \cot (\frac{π}{2}) + K = - 7$

Step 4.2.1.1.3.2

Rewrite the expression.

$π^{2} - \cot (\frac{π}{2}) + K = - 7$

Step 4.2.1.1.4

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$π^{2} - 0 + K = - 7$

Step 4.2.1.1.5

Multiply $- 1$ by $0$ .

$π^{2} + 0 + K = - 7$

Step 4.2.1.2

Add $π^{2}$ and $0$ .

$π^{2} + K = - 7$

Step 4.3

Subtract $π^{2}$ from both sides of the equation.

$K = - 7 - π^{2}$

Step 5

Substitute $- 7 - π^{2}$ for $K$ in $v = 4 t^{2} - \cot (t) + K$ and simplify.

Tap for more steps...

Step 5.1

Substitute $- 7 - π^{2}$ for $K$ .

$v = 4 t^{2} - \cot (t) - 7 - π^{2}$