Calculus Examples

$\frac{d y}{d x} = (1 + y^{2}) \tan (x)$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{1 + y^{2}}$ .

$\frac{1}{1 + y^{2}} \frac{d y}{d x} = \frac{1}{1 + y^{2}} ((1 + y^{2}) \tan (x))$

Step 1.2

Cancel the common factor of $1 + y^{2}$ .

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Step 1.2.1

Factor $1 + y^{2}$ out of $(1 + y^{2}) \tan (x)$ .

$\frac{1}{1 + y^{2}} \frac{d y}{d x} = \frac{1}{1 + y^{2}} ((1 + y^{2}) (\tan (x)))$

Step 1.2.2

Cancel the common factor.

$\frac{1}{1 + y^{2}} \frac{d y}{d x} = \frac{1}{1 + y^{2}} ((1 + y^{2}) \tan (x))$

Step 1.2.3

Rewrite the expression.

$\frac{1}{1 + y^{2}} \frac{d y}{d x} = \tan (x)$

Step 1.3

Rewrite the equation.

$\frac{1}{1 + y^{2}} d y = \tan (x) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{1 + y^{2}} d y = \int \tan (x) d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

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

$\int \frac{1}{1^{2} + y^{2}} d y = \int \tan (x) d x$

Step 2.2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

Take the inverse arctangent of both sides of the equation to extract $y$ from inside the arctangent.

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