Calculus Examples

$\frac{d v}{d t} = \frac{6}{1 + t^{2}} + \sec^{2} (t)$ , $v (0) = 3$

Step 1

Rewrite the equation.

$d v = (\frac{6}{1 + t^{2}} + \sec^{2} (t)) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d v = \int \frac{6}{1 + t^{2}} + \sec^{2} (t) d t$

Step 2.2

Apply the constant rule.

$v + C_{1} = \int \frac{6}{1 + t^{2}} + \sec^{2} (t) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$v + C_{1} = \int \frac{6}{1 + t^{2}} d t + \int \sec^{2} (t) d t$

Step 2.3.2

Since $6$ is constant with respect to $t$ , move $6$ out of the integral.

$v + C_{1} = 6 \int \frac{1}{1 + t^{2}} d t + \int \sec^{2} (t) d t$

Step 2.3.3

Rewrite $1$ as $1^{2}$ .

$v + C_{1} = 6 \int \frac{1}{1^{2} + t^{2}} d t + \int \sec^{2} (t) d t$

Step 2.3.4

The integral of $\frac{1}{1^{2} + t^{2}}$ with respect to $t$ is $\arctan (t) + C_{2}$ .

$v + C_{1} = 6 (\arctan (t) + C_{2}) + \int \sec^{2} (t) d t$

Step 2.3.5

Since the derivative of $\tan (t)$ is $\sec^{2} (t)$ , the integral of $\sec^{2} (t)$ is $\tan (t)$ .

$v + C_{1} = 6 (\arctan (t) + C_{2}) + \tan (t) + C_{3}$

Step 2.3.6

Simplify.

$v + C_{1} = 6 \arctan (t) + \tan (t) + C_{4}$

Step 2.4

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

$v = 6 \arctan (t) + \tan (t) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $t$ and $3$ for $v$ in $v = 6 \arctan (t) + \tan (t) + K$ .

$3 = 6 \arctan (0) + \tan (0) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $6 \arctan (0) + \tan (0) + K = 3$ .

$6 \arctan (0) + \tan (0) + K = 3$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $6 \arctan (0) + \tan (0) + K$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

The exact value of $\arctan (0)$ is $0$ .

$6 \cdot 0 + \tan (0) + K = 3$

Step 4.2.1.1.2

Multiply $6$ by $0$ .

$0 + \tan (0) + K = 3$

Step 4.2.1.1.3

The exact value of $\tan (0)$ is $0$ .

$0 + 0 + K = 3$

Step 4.2.1.2

Combine the opposite terms in $0 + 0 + K$ .

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Step 4.2.1.2.1

Add $0$ and $0$ .

$0 + K = 3$

Step 4.2.1.2.2

Add $0$ and $K$ .

$K = 3$

Step 5

Substitute $3$ for $K$ in $v = 6 \arctan (t) + \tan (t) + K$ and simplify.

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Step 5.1

Substitute $3$ for $K$ .

$v = 6 \arctan (t) + \tan (t) + 3$