Calculus Examples

$x \cos^{2} (y) d x + \tan (y) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = x \cos^{2} (y)$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x \cos^{2} (y)]$

Step 1.2

Since $x$ is constant with respect to $y$ , the derivative of $x \cos^{2} (y)$ with respect to $y$ is $x \frac{d}{d y} [\cos^{2} (y)]$ .

$\frac{\partial M}{\partial y} = x \frac{d}{d y} [\cos^{2} (y)]$

Step 1.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = \cos (y)$ .

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Step 1.3.1

To apply the Chain Rule, set $u$ as $\cos (y)$ .

$\frac{\partial M}{\partial y} = x (\frac{d}{d u} [u^{2}] \frac{d}{d y} [\cos (y)])$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$\frac{\partial M}{\partial y} = x (2 u \frac{d}{d y} [\cos (y)])$

Step 1.3.3

Replace all occurrences of $u$ with $\cos (y)$ .

$\frac{\partial M}{\partial y} = x (2 \cos (y) \frac{d}{d y} [\cos (y)])$

Step 1.4

Move $2$ to the left of $x$ .

$\frac{\partial M}{\partial y} = 2 \cdot x \cos (y) \frac{d}{d y} [\cos (y)]$

Step 1.5

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

$\frac{\partial M}{\partial y} = 2 x \cos (y) (- \sin (y))$

Step 1.6

Multiply $- 1$ by $2$ .

$\frac{\partial M}{\partial y} = - 2 x \cos (y) \sin (y)$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = \tan (y)$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [\tan (y)]$

Step 2.2

Since $\tan (y)$ is constant with respect to $x$ , the derivative of $\tan (y)$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 0$

Step 3

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 3.1

Substitute $- 2 x \cos (y) \sin (y)$ for $\frac{\partial M}{\partial y}$ and $0$ for $\frac{\partial N}{\partial x}$ .

$- 2 x \cos (y) \sin (y) = 0$

Step 3.2

Since the left side does not equal the right side, the equation is not an identity.

$- 2 x \cos (y) \sin (y) = 0$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

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Step 4.1

Substitute $0$ for $\frac{\partial N}{\partial x}$ .

$\frac{0 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $- 2 x \cos (y) \sin (y)$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - (- 2 x \cos (y) \sin (y))}{M}$

Step 4.3

Substitute $x \cos^{2} (y)$ for $M$ .

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Step 4.3.1

Substitute $x \cos^{2} (y)$ for $M$ .

$\frac{0 - (- 2 x \cos (y) \sin (y))}{x \cos^{2} (y)}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Multiply $- 2$ by $- 1$ .

$\frac{0 + 2 (x \cos (y) \sin (y))}{x \cos^{2} (y)}$

Step 4.3.2.2

Reorder $2$ and $x \cos (y) \sin (y)$ .

$\frac{0 + x \cos (y) \sin (y) \cdot 2}{x \cos^{2} (y)}$

Step 4.3.2.3

Add parentheses.

$\frac{0 + x \cos (y) (\sin (y) \cdot 2)}{x \cos^{2} (y)}$

Step 4.3.2.4

Add parentheses.

$\frac{0 + x (\cos (y) (\sin (y) \cdot 2))}{x \cos^{2} (y)}$

Step 4.3.2.5

Reorder $\cos (y)$ and $\sin (y) \cdot 2$ .

$\frac{0 + x (\sin (y) \cdot 2 \cos (y))}{x \cos^{2} (y)}$

Step 4.3.2.6

Reorder $\sin (y)$ and $2$ .

$\frac{0 + x (2 \cdot \sin (y) \cos (y))}{x \cos^{2} (y)}$

Step 4.3.2.7

Apply the sine double-angle identity.

$\frac{0 + x \sin (2 y)}{x \cos^{2} (y)}$

Step 4.3.2.8

Add $0$ and $x \sin (2 y)$ .

$\frac{x \sin (2 y)}{x \cos^{2} (y)}$

Step 4.3.3

Cancel the common factor of $x$ .

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Step 4.3.3.1

Cancel the common factor.

$\frac{x \sin (2 y)}{x \cos^{2} (y)}$

Step 4.3.3.2

Rewrite the expression.

$\frac{\sin (2 y)}{\cos^{2} (y)}$

Step 4.3.4

Apply the sine double-angle identity.

$\frac{2 \sin (y) \cos (y)}{\cos^{2} (y)}$

Step 4.3.5

Cancel the common factor of $\cos (y)$ and $\cos^{2} (y)$ .

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Step 4.3.5.1

Factor $\cos (y)$ out of $2 \sin (y) \cos (y)$ .

$\frac{\cos (y) (2 \sin (y))}{\cos^{2} (y)}$

Step 4.3.5.2

Cancel the common factors.

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Step 4.3.5.2.1

Factor $\cos (y)$ out of $\cos^{2} (y)$ .

$\frac{\cos (y) (2 \sin (y))}{\cos (y) \cos (y)}$

Step 4.3.5.2.2

Cancel the common factor.

$\frac{\cos (y) (2 \sin (y))}{\cos (y) \cos (y)}$

Step 4.3.5.2.3

Rewrite the expression.

$\frac{2 \sin (y)}{\cos (y)}$

Step 4.3.6

Separate fractions.

$\frac{2}{1} \cdot \frac{\sin (y)}{\cos (y)}$

Step 4.3.7

Convert from $\frac{\sin (y)}{\cos (y)}$ to $\tan (y)$ .

$\frac{2}{1} \tan (y)$

Step 4.3.8

Substitute $x \cos^{2} (y)$ for $M$ .

$2 \tan (y)$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} d y}$ .

$μ (x, y) = e^{\int 2 \tan (y) d y}$

Step 5

Evaluate the integral $e^{\int 2 \tan (y) d y}$ .

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Step 5.1

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

$μ (x, y) = e^{2 \int \tan (y) d y}$

Step 5.2

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

$μ (x, y) = e^{2 (\ln (| \sec (y) |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{2 \ln (| \sec (y) |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

Simplify $2 \ln (| \sec (y) |)$ by moving $2$ inside the logarithm.

$μ (x, y) = e^{\ln ({| \sec (y) |}^{2})} + C$

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| \sec (y) |}^{2} + C$

Step 5.4.3

Remove the absolute value in ${| \sec (y) |}^{2}$ because exponentiations with even powers are always positive.

$μ (x, y) = \sec^{2} (y) + C$

Step 6

Multiply both sides of $x \cos^{2} (y) d x + \tan (y) d y = 0$ by the integration factor $\sec^{2} (y)$ .

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Step 6.1

Multiply $x \cos^{2} (y)$ by $\sec^{2} (y)$ .

$x \cos^{2} (y) \sec^{2} (y) d x + \tan (y) d y = 0$

Step 6.2

Rewrite $\sec (y)$ in terms of sines and cosines.

$x \cos^{2} (y) {(\frac{1}{\cos (y)})}^{2} d x + \tan (y) d y = 0$

Step 6.3

Apply the product rule to $\frac{1}{\cos (y)}$ .

$x \cos^{2} (y) \frac{1^{2}}{\cos^{2} (y)} d x + \tan (y) d y = 0$

Step 6.4

Cancel the common factor of $\cos^{2} (y)$ .

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Step 6.4.1

Factor $\cos^{2} (y)$ out of $x \cos^{2} (y)$ .

$\cos^{2} (y) x \frac{1^{2}}{\cos^{2} (y)} d x + \tan (y) d y = 0$

Step 6.4.2

Cancel the common factor.

$\cos^{2} (y) x \frac{1^{2}}{\cos^{2} (y)} d x + \tan (y) d y = 0$

Step 6.4.3

Rewrite the expression.

$x \cdot 1^{2} d x + \tan (y) d y = 0$

Step 6.5

One to any power is one.

$x \cdot 1 d x + \tan (y) d y = 0$

Step 6.6

Multiply $x$ by $1$ .

$x d x + \tan (y) d y = 0$

Step 6.7

Multiply $\tan (y)$ by $\sec^{2} (y)$ .

$x d x + \tan (y) \sec^{2} (y) d y = 0$

Step 7

Set $f (x, y)$ equal to the integral of $M (x, y)$ .

$f (x, y) = \int x d x$

Step 8

Integrate $M (x, y) = x$ to find $f (x, y)$ .

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Step 8.1

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

$f (x, y) = \frac{1}{2} x^{2} + C$

Step 9

Since the integral of $g (y)$ will contain an integration constant, we can replace $C$ with $g (y)$ .

$f (x, y) = \frac{1}{2} x^{2} + g (y)$

Step 10

Set $\frac{\partial f}{\partial y} = N (x, y)$ .

$\frac{\partial f}{\partial y} = \tan (y) \sec^{2} (y)$

Step 11

Find $\frac{\partial f}{\partial y}$ .

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Step 11.1

Differentiate $f$ with respect to $y$ .

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

Step 11.2

By the Sum Rule, the derivative of $\frac{1}{2} x^{2} + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{2} x^{2}] + \frac{d}{d y} [g (y)]$ .

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

Step 11.3

Since $\frac{1}{2} x^{2}$ is constant with respect to $y$ , the derivative of $\frac{1}{2} x^{2}$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [g (y)] = \tan (y) \sec^{2} (y)$

Step 11.4

Differentiate using the function rule which states that the derivative of $g (y)$ is $\frac{d g}{d y}$ .

$0 + \frac{d g}{d y} = \tan (y) \sec^{2} (y)$

Step 11.5

Add $0$ and $\frac{d g}{d y}$ .

$\frac{d g}{d y} = \tan (y) \sec^{2} (y)$

Step 12

Find the antiderivative of $\tan (y) \sec^{2} (y)$ to find $g (y)$ .

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Step 12.1

Integrate both sides of $\frac{d g}{d y} = \tan (y) \sec^{2} (y)$ .

$\int \frac{d g}{d y} d y = \int \tan (y) \sec^{2} (y) d y$

Step 12.2

Evaluate $\int \frac{d g}{d y} d y$ .

$g (y) = \int \tan (y) \sec^{2} (y) d y$

Step 12.3

Let $u = \sec (y)$ . Then $d u = \sec (y) \tan (y) d y$ , so $\frac{1}{\sec (y) \tan (y)} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 12.3.1

Let $u = \sec (y)$ . Find $\frac{d u}{d y}$ .

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Step 12.3.1.1

Differentiate $\sec (y)$ .

$\frac{d}{d y} [\sec (y)]$

Step 12.3.1.2

The derivative of $\sec (y)$ with respect to $y$ is $\sec (y) \tan (y)$ .

$\sec (y) \tan (y)$

Step 12.3.2

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

$g (y) = \int u d u$

Step 12.4

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

$g (y) = \frac{1}{2} u^{2} + C$

Step 12.5

Replace all occurrences of $u$ with $\sec (y)$ .

$g (y) = \frac{1}{2} \sec^{2} (y) + C$

Step 13

Substitute for $g (y)$ in $f (x, y) = \frac{1}{2} x^{2} + g (y)$ .

$f (x, y) = \frac{1}{2} x^{2} + \frac{1}{2} \sec^{2} (y) + C$

Step 14

Simplify each term.

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Step 14.1

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

$f (x, y) = \frac{x^{2}}{2} + \frac{1}{2} \sec^{2} (y) + C$

Step 14.2

Combine $\frac{1}{2}$ and $\sec^{2} (y)$ .

$f (x, y) = \frac{x^{2}}{2} + \frac{\sec^{2} (y)}{2} + C$