Calculus Examples

$\frac{d y}{d x} = \sec^{2} (x) {(2 + y)}^{2}$ , $y (π) = - 5$

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

Cancel the common factor.

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

Step 1.2.3

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 2 + y$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 2 + y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $2 + y$ .

$\frac{d}{d y} [2 + y]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $2 + y$ with respect to $y$ is $\frac{d}{d y} [2] + \frac{d}{d y} [y]$ .

$\frac{d}{d y} [2] + \frac{d}{d y} [y]$

Step 2.2.1.1.3

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y]$

Step 2.2.1.1.4

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

$0 + 1$

Step 2.2.1.1.5

Add $0$ and $1$ .

$1$

Step 2.2.1.2

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

$\int \frac{1}{u^{2}} d u = \int \sec^{2} (x) d x$

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

Move $u^{2}$ out of the denominator by raising it to the $- 1$ power.

$\int {(u^{2})}^{- 1} d u = \int \sec^{2} (x) d x$

Step 2.2.2.2

Multiply the exponents in ${(u^{2})}^{- 1}$ .

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Step 2.2.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int u^{2 \cdot - 1} d u = \int \sec^{2} (x) d x$

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

$\int u^{- 2} d u = \int \sec^{2} (x) d x$

Step 2.2.3

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

$- u^{- 1} + C_{1} = \int \sec^{2} (x) d x$

Step 2.2.4

Rewrite $- u^{- 1} + C_{1}$ as $- \frac{1}{u} + C_{1}$ .

$- \frac{1}{u} + C_{1} = \int \sec^{2} (x) d x$

Step 2.2.5

Replace all occurrences of $u$ with $2 + y$ .

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

Step 2.3

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

$- \frac{1}{2 + y} + C_{1} = \tan (x) + C_{2}$

Step 2.4

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

$- \frac{1}{2 + y} = \tan (x) + K$

Step 3

Solve for $y$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 + y, 1, 1$

Step 3.1.2

Remove parentheses.

$2 + y, 1, 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$2 + y$

Step 3.2

Multiply each term in $- \frac{1}{2 + y} = \tan (x) + K$ by $2 + y$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $- \frac{1}{2 + y} = \tan (x) + K$ by $2 + y$ .

$- \frac{1}{2 + y} (2 + y) = \tan (x) (2 + y) + K (2 + y)$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2 + y$ .

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Step 3.2.2.1.1

Move the leading negative in $- \frac{1}{2 + y}$ into the numerator.

$\frac{- 1}{2 + y} (2 + y) = \tan (x) (2 + y) + K (2 + y)$

Step 3.2.2.1.2

Cancel the common factor.

$\frac{- 1}{2 + y} (2 + y) = \tan (x) (2 + y) + K (2 + y)$

Step 3.2.2.1.3

Rewrite the expression.

$- 1 = \tan (x) (2 + y) + K (2 + y)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Apply the distributive property.

$- 1 = \tan (x) \cdot 2 + \tan (x) y + K (2 + y)$

Step 3.2.3.1.2

Move $2$ to the left of $\tan (x)$ .

$- 1 = 2 \cdot \tan (x) + \tan (x) y + K (2 + y)$

Step 3.2.3.1.3

Apply the distributive property.

$- 1 = 2 \tan (x) + \tan (x) y + K \cdot 2 + K y$

Step 3.2.3.1.4

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

$- 1 = 2 \tan (x) + \tan (x) y + 2 K + K y$

Step 3.2.3.2

Reorder factors in $2 \tan (x) + \tan (x) y + 2 K + K y$ .

$- 1 = 2 \tan (x) + y \tan (x) + 2 K + K y$

Step 3.3

Solve the equation.

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Step 3.3.1

Rewrite the equation as $2 \tan (x) + y \tan (x) + 2 K + K y = - 1$ .

$2 \tan (x) + y \tan (x) + 2 K + K y = - 1$

Step 3.3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 3.3.2.1

Subtract $2 \tan (x)$ from both sides of the equation.

$y \tan (x) + 2 K + K y = - 1 - 2 \tan (x)$

Step 3.3.2.2

Subtract $2 K$ from both sides of the equation.

$y \tan (x) + K y = - 1 - 2 \tan (x) - 2 K$

Step 3.3.3

Factor $y$ out of $y \tan (x) + K y$ .

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Step 3.3.3.1

Factor $y$ out of $y \tan (x)$ .

$y (\tan (x)) + K y = - 1 - 2 \tan (x) - 2 K$

Step 3.3.3.2

Factor $y$ out of $K y$ .

$y (\tan (x)) + y K = - 1 - 2 \tan (x) - 2 K$

Step 3.3.3.3

Factor $y$ out of $y (\tan (x)) + y K$ .

$y (\tan (x) + K) = - 1 - 2 \tan (x) - 2 K$

Step 3.3.4

Divide each term in $y (\tan (x) + K) = - 1 - 2 \tan (x) - 2 K$ by $\tan (x) + K$ and simplify.

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Step 3.3.4.1

Divide each term in $y (\tan (x) + K) = - 1 - 2 \tan (x) - 2 K$ by $\tan (x) + K$ .

$\frac{y (\tan (x) + K)}{\tan (x) + K} = \frac{- 1}{\tan (x) + K} + \frac{- 2 \tan (x)}{\tan (x) + K} + \frac{- 2 K}{\tan (x) + K}$

Step 3.3.4.2

Simplify the left side.

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Step 3.3.4.2.1

Cancel the common factor of $\tan (x) + K$ .

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Step 3.3.4.2.1.1

Cancel the common factor.

$\frac{y (\tan (x) + K)}{\tan (x) + K} = \frac{- 1}{\tan (x) + K} + \frac{- 2 \tan (x)}{\tan (x) + K} + \frac{- 2 K}{\tan (x) + K}$

Step 3.3.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 1}{\tan (x) + K} + \frac{- 2 \tan (x)}{\tan (x) + K} + \frac{- 2 K}{\tan (x) + K}$

Step 3.3.4.3

Simplify the right side.

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Step 3.3.4.3.1

Simplify terms.

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Step 3.3.4.3.1.1

Simplify each term.

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Step 3.3.4.3.1.1.1

Move the negative in front of the fraction.

$y = - \frac{1}{\tan (x) + K} + \frac{- 2 \tan (x)}{\tan (x) + K} + \frac{- 2 K}{\tan (x) + K}$

Step 3.3.4.3.1.1.2

Move the negative in front of the fraction.

$y = - \frac{1}{\tan (x) + K} - \frac{2 \tan (x)}{\tan (x) + K} + \frac{- 2 K}{\tan (x) + K}$

Step 3.3.4.3.1.1.3

Move the negative in front of the fraction.

$y = - \frac{1}{\tan (x) + K} - \frac{2 \tan (x)}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.2

Combine the numerators over the common denominator.

$y = \frac{- 1 - 2 \tan (x)}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.3

Simplify each term.

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Step 3.3.4.3.1.3.1

Factor $- 1$ out of $- 1 - 2 \tan (x)$ .

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Step 3.3.4.3.1.3.1.1

Rewrite $- 1$ as $- 1 (1)$ .

$y = \frac{- 1 (1) - 2 \tan (x)}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.3.1.2

Factor $- 1$ out of $- 2 \tan (x)$ .

$y = \frac{- 1 (1) - (2 \tan (x))}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.3.1.3

Factor $- 1$ out of $- 1 (1) - (2 \tan (x))$ .

$y = \frac{- 1 (1 + 2 \tan (x))}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.3.1.4

Rewrite $- 1 (1 + 2 \tan (x))$ as $- (1 + 2 \tan (x))$ .

$y = \frac{- (1 + 2 \tan (x))}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.3.2

Move the negative in front of the fraction.

$y = - \frac{1 + 2 \tan (x)}{\tan (x) + K} - \frac{2 K}{\tan (x) + K}$

Step 3.3.4.3.1.4

Combine the numerators over the common denominator.

$y = \frac{- (1 + 2 \tan (x)) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.2

Simplify the numerator.

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Step 3.3.4.3.2.1

Apply the distributive property.

$y = \frac{- 1 \cdot 1 - (2 \tan (x)) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.2.2

Multiply $- 1$ by $1$ .

$y = \frac{- 1 - (2 \tan (x)) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.2.3

Multiply $2$ by $- 1$ .

$y = \frac{- 1 - 2 \tan (x) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.3

Simplify with factoring out.

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Step 3.3.4.3.3.1

Rewrite $- 1$ as $- 1 (1)$ .

$y = \frac{- 1 (1) - 2 \tan (x) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.3.2

Factor $- 1$ out of $- 2 \tan (x)$ .

$y = \frac{- 1 (1) - (2 \tan (x)) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.3.3

Factor $- 1$ out of $- 1 (1) - (2 \tan (x))$ .

$y = \frac{- 1 (1 + 2 \tan (x)) - 2 K}{\tan (x) + K}$

Step 3.3.4.3.3.4

Factor $- 1$ out of $- 2 K$ .

$y = \frac{- 1 (1 + 2 \tan (x)) - (2 K)}{\tan (x) + K}$

Step 3.3.4.3.3.5

Factor $- 1$ out of $- 1 (1 + 2 \tan (x)) - (2 K)$ .

$y = \frac{- 1 (1 + 2 \tan (x) + 2 K)}{\tan (x) + K}$

Step 3.3.4.3.3.6

Move the negative in front of the fraction.

$y = - \frac{1 + 2 \tan (x) + 2 K}{\tan (x) + K}$

Step 4

Simplify the constant of integration.

$y = - \frac{1 + 2 \tan (x) + K}{\tan (x) + K}$

Step 5

Use the initial condition to find the value of $K$ by substituting $π$ for $x$ and $- 5$ for $y$ in $y = - \frac{1 + 2 \tan (x) + K}{\tan (x) + K}$ .

$- 5 = - \frac{1 + 2 \tan (π) + K}{\tan (π) + K}$

Step 6

Solve for $K$ .

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Step 6.1

Rewrite the equation as $- \frac{1 + 2 \tan (π) + K}{\tan (π) + K} = - 5$ .

$- \frac{1 + 2 \tan (π) + K}{\tan (π) + K} = - 5$

Step 6.2

Factor each term.

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Step 6.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$- \frac{1 + 2 (- \tan (0)) + K}{\tan (π) + K} = - 5$

Step 6.2.2

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

$- \frac{1 + 2 (- 0) + K}{\tan (π) + K} = - 5$

Step 6.2.3

Multiply $2 (- 0)$ .

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Step 6.2.3.1

Multiply $- 1$ by $0$ .

$- \frac{1 + 2 \cdot 0 + K}{\tan (π) + K} = - 5$

Step 6.2.3.2

Multiply $2$ by $0$ .

$- \frac{1 + 0 + K}{\tan (π) + K} = - 5$

Step 6.2.4

Add $1$ and $0$ .

$- \frac{1 + K}{\tan (π) + K} = - 5$

Step 6.2.5

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$- \frac{1 + K}{- \tan (0) + K} = - 5$

Step 6.2.6

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

$- \frac{1 + K}{- 0 + K} = - 5$

Step 6.2.7

Multiply $- 1$ by $0$ .

$- \frac{1 + K}{0 + K} = - 5$

Step 6.2.8

Add $0$ and $K$ .

$- \frac{1 + K}{K} = - 5$

Step 6.3

Find the LCD of the terms in the equation.

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Step 6.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$K, 1$

Step 6.3.2

The LCM of one and any expression is the expression.

$K$

Step 6.4

Multiply each term in $- \frac{1 + K}{K} = - 5$ by $K$ to eliminate the fractions.

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Step 6.4.1

Multiply each term in $- \frac{1 + K}{K} = - 5$ by $K$ .

$- \frac{1 + K}{K} K = - 5 K$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Cancel the common factor of $K$ .

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Step 6.4.2.1.1

Move the leading negative in $- \frac{1 + K}{K}$ into the numerator.

$\frac{- (1 + K)}{K} K = - 5 K$

Step 6.4.2.1.2

Cancel the common factor.

$\frac{- (1 + K)}{K} K = - 5 K$

Step 6.4.2.1.3

Rewrite the expression.

$- (1 + K) = - 5 K$

Step 6.4.2.2

Apply the distributive property.

$- 1 \cdot 1 - K = - 5 K$

Step 6.4.2.3

Multiply $- 1$ by $1$ .

$- 1 - K = - 5 K$

Step 6.5

Solve the equation.

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Step 6.5.1

Move all terms containing $K$ to the left side of the equation.

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Step 6.5.1.1

Add $5 K$ to both sides of the equation.

$- 1 - K + 5 K = 0$

Step 6.5.1.2

Add $- K$ and $5 K$ .

$- 1 + 4 K = 0$

Step 6.5.2

Add $1$ to both sides of the equation.

$4 K = 1$

Step 6.5.3

Divide each term in $4 K = 1$ by $4$ and simplify.

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Step 6.5.3.1

Divide each term in $4 K = 1$ by $4$ .

$\frac{4 K}{4} = \frac{1}{4}$

Step 6.5.3.2

Simplify the left side.

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Step 6.5.3.2.1

Cancel the common factor of $4$ .

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Step 6.5.3.2.1.1

Cancel the common factor.

$\frac{4 K}{4} = \frac{1}{4}$

Step 6.5.3.2.1.2

Divide $K$ by $1$ .

$K = \frac{1}{4}$

Step 7

Substitute $\frac{1}{4}$ for $K$ in $y = - \frac{1 + 2 \tan (x) + K}{\tan (x) + K}$ and simplify.

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Step 7.1

Substitute $\frac{1}{4}$ for $K$ .

$y = - \frac{1 + 2 \tan (x) + \frac{1}{4}}{\tan (x) + K}$

Step 7.2

Multiply the numerator and denominator of the fraction by $4$ .

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Step 7.2.1

Multiply $\frac{1 + 2 \tan (x) + \frac{1}{4}}{\tan (x) + K}$ by $\frac{4}{4}$ .

$y = - (\frac{4}{4} \cdot \frac{1 + 2 \tan (x) + \frac{1}{4}}{\tan (x) + K})$

Step 7.2.2

Combine.

$y = - \frac{4 (1 + 2 \tan (x) + \frac{1}{4})}{4 (\tan (x) + K)}$

Step 7.3

Apply the distributive property.

$y = - \frac{4 \cdot 1 + 4 (2 \tan (x)) + 4 (\frac{1}{4})}{4 \tan (x) + 4 K}$

Step 7.4

Cancel the common factor of $4$ .

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Step 7.4.1

Cancel the common factor.

$y = - \frac{4 \cdot 1 + 4 (2 \tan (x)) + 4 (\frac{1}{4})}{4 \tan (x) + 4 K}$

Step 7.4.2

Rewrite the expression.

$y = - \frac{4 \cdot 1 + 4 (2 \tan (x)) + 1}{4 \tan (x) + 4 K}$

Step 7.5

Simplify the numerator.

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Step 7.5.1

Multiply $4$ by $1$ .

$y = - \frac{4 + 4 \cdot 2 \tan (x) + 1}{4 \tan (x) + 4 K}$

Step 7.5.2

Multiply $4$ by $2$ .

$y = - \frac{4 + 8 \tan (x) + 1}{4 \tan (x) + 4 K}$

Step 7.5.3

Add $4$ and $1$ .

$y = - \frac{5 + 8 \tan (x)}{4 \tan (x) + 4 K}$

Step 7.6

Factor $4$ out of $4 \tan (x) + 4 K$ .

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Step 7.6.1

Factor $4$ out of $4 \tan (x)$ .

$y = - \frac{5 + 8 \tan (x)}{4 (\tan (x)) + 4 K}$

Step 7.6.2

Factor $4$ out of $4 K$ .

$y = - \frac{5 + 8 \tan (x)}{4 (\tan (x)) + 4 (K)}$

Step 7.6.3

Factor $4$ out of $4 (\tan (x)) + 4 (K)$ .

$y = - \frac{5 + 8 \tan (x)}{4 (\tan (x) + K)}$