Calculus Examples

$2 y \frac{d y}{d x} = \frac{\cos (x)}{1 + \sin (x)}$

Step 1

Rewrite the equation.

$2 y d y = \frac{\cos (x)}{1 + \sin (x)} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int 2 y d y = \int \frac{\cos (x)}{1 + \sin (x)} d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Since $2$ is constant with respect to $y$ , move $2$ out of the integral.

$2 \int y d y = \int \frac{\cos (x)}{1 + \sin (x)} d x$

Step 2.2.2

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

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

Step 2.2.3

Simplify the answer.

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Step 2.2.3.1

Rewrite $2 (\frac{1}{2} y^{2} + C_{1})$ as $2 (\frac{1}{2}) y^{2} + C_{1}$ .

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

Step 2.2.3.2

Simplify.

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Step 2.2.3.2.1

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

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

Step 2.2.3.2.2

Cancel the common factor of $2$ .

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Step 2.2.3.2.2.1

Cancel the common factor.

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

Step 2.2.3.2.2.2

Rewrite the expression.

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

Step 2.2.3.2.3

Multiply $y^{2}$ by $1$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = 1 + \sin (x)$ . Then $d u = \cos (x) d x$ , so $\frac{1}{\cos (x)} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = 1 + \sin (x)$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $1 + \sin (x)$ .

$\frac{d}{d x} [1 + \sin (x)]$

Step 2.3.1.1.2

Differentiate.

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Step 2.3.1.1.2.1

By the Sum Rule, the derivative of $1 + \sin (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\sin (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\sin (x)]$

Step 2.3.1.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\sin (x)]$

Step 2.3.1.1.3

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$0 + \cos (x)$

Step 2.3.1.1.4

Add $0$ and $\cos (x)$ .

$\cos (x)$

Step 2.3.1.2

Rewrite the problem using $u$ and $d u$ .

$y^{2} + C_{1} = \int \frac{1}{u} d u$

Step 2.3.2

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$y^{2} + C_{1} = \ln (| u |) + C_{2}$

Step 2.3.3

Replace all occurrences of $u$ with $1 + \sin (x)$ .

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

Step 2.4

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

$y^{2} = \ln (| 1 + \sin (x) |) + C$

Step 3

Solve for $y$ .

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Step 3.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\ln (| 1 + \sin (x) |) + C}$

Step 3.2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.2.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{\ln (| 1 + \sin (x) |) + C}$

Step 3.2.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{\ln (| 1 + \sin (x) |) + C}$

Step 3.2.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{\ln (| 1 + \sin (x) |) + C}$

$y = - \sqrt{\ln (| 1 + \sin (x) |) + C}$

$y = \sqrt{\ln (| 1 + \sin (x) |) + C}$

$y = - \sqrt{\ln (| 1 + \sin (x) |) + C}$

$y = \sqrt{\ln (| 1 + \sin (x) |) + C}$

$y = - \sqrt{\ln (| 1 + \sin (x) |) + C}$