Calculus Examples

$\frac{d y}{d θ} = 4 y^{2} \sec^{2} (2 θ)$

Step 1

Separate the variables.

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Step 1.1

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

$\frac{1}{y^{2}} \frac{d y}{d θ} = \frac{1}{y^{2}} (4 y^{2} \sec^{2} (2 θ))$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{y^{2}} \frac{d y}{d θ} = 4 \frac{1}{y^{2}} (y^{2} \sec^{2} (2 θ))$

Step 1.2.2

Combine $4$ and $\frac{1}{y^{2}}$ .

$\frac{1}{y^{2}} \frac{d y}{d θ} = \frac{4}{y^{2}} (y^{2} \sec^{2} (2 θ))$

Step 1.2.3

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

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Step 1.2.3.1

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

$\frac{1}{y^{2}} \frac{d y}{d θ} = \frac{4}{y^{2}} (y^{2} (\sec^{2} (2 θ)))$

Step 1.2.3.2

Cancel the common factor.

$\frac{1}{y^{2}} \frac{d y}{d θ} = \frac{4}{y^{2}} (y^{2} \sec^{2} (2 θ))$

Step 1.2.3.3

Rewrite the expression.

$\frac{1}{y^{2}} \frac{d y}{d θ} = 4 \sec^{2} (2 θ)$

Step 1.3

Rewrite the equation.

$\frac{1}{y^{2}} d y = 4 \sec^{2} (2 θ) d θ$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y^{2}} d y = \int 4 \sec^{2} (2 θ) d θ$

Step 2.2

Integrate the left side.

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Step 2.2.1

Apply basic rules of exponents.

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Step 2.2.1.1

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

$\int {(y^{2})}^{- 1} d y = \int 4 \sec^{2} (2 θ) d θ$

Step 2.2.1.2

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

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Step 2.2.1.2.1

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

$\int y^{2 \cdot - 1} d y = \int 4 \sec^{2} (2 θ) d θ$

Step 2.2.1.2.2

Multiply $2$ by $- 1$ .

$\int y^{- 2} d y = \int 4 \sec^{2} (2 θ) d θ$

Step 2.2.2

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

$- y^{- 1} + C_{1} = \int 4 \sec^{2} (2 θ) d θ$

Step 2.2.3

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

$- \frac{1}{y} + C_{1} = \int 4 \sec^{2} (2 θ) d θ$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $4$ is constant with respect to $θ$ , move $4$ out of the integral.

$- \frac{1}{y} + C_{1} = 4 \int \sec^{2} (2 θ) d θ$

Step 2.3.2

Let $u = 2 θ$ . Then $d u = 2 d θ$ , so $\frac{1}{2} d u = d θ$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

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

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Step 2.3.2.1.1

Differentiate $2 θ$ .

$\frac{d}{d θ} [2 θ]$

Step 2.3.2.1.2

Since $2$ is constant with respect to $θ$ , the derivative of $2 θ$ with respect to $θ$ is $2 \frac{d}{d θ} [θ]$ .

$2 \frac{d}{d θ} [θ]$

Step 2.3.2.1.3

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

$2 \cdot 1$

Step 2.3.2.1.4

Multiply $2$ by $1$ .

$2$

Step 2.3.2.2

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

$- \frac{1}{y} + C_{1} = 4 \int \sec^{2} (u) \frac{1}{2} d u$

Step 2.3.3

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

$- \frac{1}{y} + C_{1} = 4 \int \frac{\sec^{2} (u)}{2} d u$

Step 2.3.4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{y} + C_{1} = 4 (\frac{1}{2} \int \sec^{2} (u) d u)$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{2}$ and $4$ .

$- \frac{1}{y} + C_{1} = \frac{4}{2} \int \sec^{2} (u) d u$

Step 2.3.5.2

Cancel the common factor of $4$ and $2$ .

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Step 2.3.5.2.1

Factor $2$ out of $4$ .

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

Step 2.3.5.2.2

Cancel the common factors.

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Step 2.3.5.2.2.1

Factor $2$ out of $2$ .

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

Step 2.3.5.2.2.2

Cancel the common factor.

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

Step 2.3.5.2.2.3

Rewrite the expression.

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

Step 2.3.5.2.2.4

Divide $2$ by $1$ .

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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

Step 2.3.8

Replace all occurrences of $u$ with $2 θ$ .

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

Step 2.4

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

$- \frac{1}{y} = 2 \tan (2 θ) + K$

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.

$y, 1, 1$

Step 3.1.2

The LCM of one and any expression is the expression.

$y$

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $y$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Cancel the common factor.

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

Step 3.2.2.1.3

Rewrite the expression.

$- 1 = 2 \tan (2 θ) y + K y$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

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

$- 1 = 2 y \tan (2 θ) + K y$

Step 3.3

Solve the equation.

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Step 3.3.1

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

$2 y \tan (2 θ) + K y = - 1$

Step 3.3.2

Factor $y$ out of $2 y \tan (2 θ) + K y$ .

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Step 3.3.2.1

Factor $y$ out of $2 y \tan (2 θ)$ .

$y (2 \tan (2 θ)) + K y = - 1$

Step 3.3.2.2

Factor $y$ out of $K y$ .

$y (2 \tan (2 θ)) + y K = - 1$

Step 3.3.2.3

Factor $y$ out of $y (2 \tan (2 θ)) + y K$ .

$y (2 \tan (2 θ) + K) = - 1$

Step 3.3.3

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

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Step 3.3.3.1

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

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

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $2 \tan (2 θ) + K$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

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

Step 3.3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 1}{2 \tan (2 θ) + K}$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Move the negative in front of the fraction.

$y = - \frac{1}{2 \tan (2 θ) + K}$