Calculus Examples

$\frac{d y}{d x} = \tan (x)$ , $y (0) = 1$

Step 1

Rewrite the equation.

$d y = \tan (x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \tan (x) d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \tan (x) d x$

Step 2.3

The integral of $\tan (x)$ with respect to $x$ is $\ln (| \sec (x) |)$ .

$y + C_{1} = \ln (| \sec (x) |) + C_{2}$

Step 2.4

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

$y = \ln (| \sec (x) |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $1$ for $y$ in $y = \ln (| \sec (x) |) + C$ .

$1 = \ln (| \sec (0) |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $\ln (| \sec (0) |) + C = 1$ .

$\ln (| \sec (0) |) + C = 1$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $\ln (| \sec (0) |) + C$ .

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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 $\sec (0)$ is $1$ .

$\ln (| 1 |) + C = 1$

Step 4.2.1.1.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\ln (1) + C = 1$

Step 4.2.1.1.3

The natural logarithm of $1$ is $0$ .

$0 + C = 1$

Step 4.2.1.2

Add $0$ and $C$ .

$C = 1$

Step 5

Substitute $1$ for $C$ in $y = \ln (| \sec (x) |) + C$ and simplify.

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Step 5.1

Substitute $1$ for $C$ .

$y = \ln (| \sec (x) |) + 1$