Calculus Examples

$(1 + \sin^{2} (x)) \frac{d y}{d x} = e^{- 2 y} \sin (2 x)$ , $y (0) = 1$

Step 1

Separate the variables.

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Step 1.1

Divide each term in $(1 + \sin^{2} (x)) \frac{d y}{d x} = e^{- 2 y} \sin (2 x)$ by $1 + \sin^{2} (x)$ and simplify.

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Step 1.1.1

Divide each term in $(1 + \sin^{2} (x)) \frac{d y}{d x} = e^{- 2 y} \sin (2 x)$ by $1 + \sin^{2} (x)$ .

$\frac{(1 + \sin^{2} (x)) \frac{d y}{d x}}{1 + \sin^{2} (x)} = \frac{e^{- 2 y} \sin (2 x)}{1 + \sin^{2} (x)}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $1 + \sin^{2} (x)$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{(1 + \sin^{2} (x)) \frac{d y}{d x}}{1 + \sin^{2} (x)} = \frac{e^{- 2 y} \sin (2 x)}{1 + \sin^{2} (x)}$

Step 1.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{e^{- 2 y} \sin (2 x)}{1 + \sin^{2} (x)}$

Step 1.2

Regroup factors.

$\frac{d y}{d x} = e^{- 2 y} \frac{\sin (2 x)}{1 + \sin^{2} (x)}$

Step 1.3

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

$\frac{1}{e^{- 2 y}} \frac{d y}{d x} = \frac{1}{e^{- 2 y}} (e^{- 2 y} \frac{\sin (2 x)}{1 + \sin^{2} (x)})$

Step 1.4

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

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Step 1.4.1

Cancel the common factor.

$\frac{1}{e^{- 2 y}} \frac{d y}{d x} = \frac{1}{e^{- 2 y}} (e^{- 2 y} \frac{\sin (2 x)}{1 + \sin^{2} (x)})$

Step 1.4.2

Rewrite the expression.

$\frac{1}{e^{- 2 y}} \frac{d y}{d x} = \frac{\sin (2 x)}{1 + \sin^{2} (x)}$

Step 1.5

Rewrite the equation.

$\frac{1}{e^{- 2 y}} d y = \frac{\sin (2 x)}{1 + \sin^{2} (x)} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Negate the exponent of $e^{- 2 y}$ and move it out of the denominator.

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

Step 2.2.1.2

Simplify.

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Step 2.2.1.2.1

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

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Step 2.2.1.2.1.1

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

$\int 1 e^{- 2 y \cdot - 1} d y = \int \frac{\sin (2 x)}{1 + \sin^{2} (x)} d x$

Step 2.2.1.2.1.2

Multiply $- 1$ by $- 2$ .

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

Step 2.2.1.2.2

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

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

Step 2.2.2

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

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Step 2.2.2.1

Let $u_{1} = 2 y$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.2.1.1

Differentiate $2 y$ .

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

Step 2.2.2.1.2

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

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

Step 2.2.2.1.3

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

$2 \cdot 1$

Step 2.2.2.1.4

Multiply $2$ by $1$ .

$2$

Step 2.2.2.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int e^{u_{1}} \frac{1}{2} d u_{1} = \int \frac{\sin (2 x)}{1 + \sin^{2} (x)} d x$

Step 2.2.3

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

$\int \frac{e^{u_{1}}}{2} d u_{1} = \int \frac{\sin (2 x)}{1 + \sin^{2} (x)} d x$

Step 2.2.4

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

$\frac{1}{2} \int e^{u_{1}} d u_{1} = \int \frac{\sin (2 x)}{1 + \sin^{2} (x)} d x$

Step 2.2.5

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Replace all occurrences of $u_{1}$ with $2 y$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{3} = 1 + \sin^{2} (x)$ . Then $d u_{3} = \sin (2 x) d x$ , so $\frac{1}{\sin (2 x)} d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 2.3.1.1

Let $u_{3} = 1 + \sin^{2} (x)$ . Find $\frac{d u_{3}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $1 + \sin^{2} (x)$ .

$\frac{d}{d x} [1 + \sin^{2} (x)]$

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 $1 + \sin^{2} (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\sin^{2} (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\sin^{2} (x)]$

Step 2.3.1.1.2.2

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

$0 + \frac{d}{d x} [\sin^{2} (x)]$

Step 2.3.1.1.3

Evaluate $\frac{d}{d x} [\sin^{2} (x)]$ .

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Step 2.3.1.1.3.1

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

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Step 2.3.1.1.3.1.1

To apply the Chain Rule, set $u_{2}$ as $\sin (x)$ .

$0 + \frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [\sin (x)]$

Step 2.3.1.1.3.1.2

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

$0 + 2 u_{2} \frac{d}{d x} [\sin (x)]$

Step 2.3.1.1.3.1.3

Replace all occurrences of $u_{2}$ with $\sin (x)$ .

$0 + 2 \sin (x) \frac{d}{d x} [\sin (x)]$

Step 2.3.1.1.3.2

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

$0 + 2 \sin (x) \cos (x)$

Step 2.3.1.1.4

Simplify.

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Step 2.3.1.1.4.1

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

$2 \sin (x) \cos (x)$

Step 2.3.1.1.4.2

Reorder the factors of $2 \sin (x) \cos (x)$ .

$2 \cos (x) \sin (x)$

Step 2.3.1.1.4.3

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

$\sin (x) (2 \cos (x))$

Step 2.3.1.1.4.4

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

$2 \cdot \sin (x) \cos (x)$

Step 2.3.1.1.4.5

Apply the sine double-angle identity.

$\sin (2 x)$

Step 2.3.1.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$\frac{1}{2} e^{2 y} + C_{1} = \int \frac{1}{u_{3}} d u_{3}$

Step 2.3.2

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$\frac{1}{2} e^{2 y} + C_{1} = \ln (| u_{3} |) + C_{2}$

Step 2.3.3

Replace all occurrences of $u_{3}$ with $1 + \sin^{2} (x)$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $2 (\frac{1}{2} e^{2 y})$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

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

Step 3.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 3.4

Expand the left side.

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Step 3.4.1

Expand $\ln (e^{2 y})$ by moving $2 y$ outside the logarithm.

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

Step 3.4.2

The natural logarithm of $e$ is $1$ .

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

Step 3.4.3

Multiply $2$ by $1$ .

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

Step 3.5

Expand $\ln ({(1 + \sin^{2} (x))}^{2})$ by moving $2$ outside the logarithm.

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

Step 3.6

Divide each term in $2 y = \ln (2 \ln (1 + \sin^{2} (x)) + 2 C)$ by $2$ and simplify.

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Step 3.6.1

Divide each term in $2 y = \ln (2 \ln (1 + \sin^{2} (x)) + 2 C)$ by $2$ .

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

Step 3.6.2

Simplify the left side.

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Step 3.6.2.1

Cancel the common factor of $2$ .

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Step 3.6.2.1.1

Cancel the common factor.

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

Step 3.6.2.1.2

Divide $y$ by $1$ .

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

Step 4

Simplify the constant of integration.

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

Step 5

Use the initial condition to find the value of $C$ by substituting $0$ for $x$ and $1$ for $y$ in $y = \frac{\ln (2 \ln (1 + \sin^{2} (x)) + C)}{2}$ .

$1 = \frac{\ln (2 \ln (1 + \sin^{2} (0)) + C)}{2}$

Step 6

Solve for $C$ .

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Step 6.1

Rewrite the equation as $\frac{\ln (2 \ln (1 + \sin^{2} (0)) + C)}{2} = 1$ .

$\frac{\ln (2 \ln (1 + \sin^{2} (0)) + C)}{2} = 1$

Step 6.2

Multiply both sides of the equation by $2$ .

$2 \frac{\ln (2 \ln (1 + \sin^{2} (0)) + C)}{2} = 2 \cdot 1$

Step 6.3

Simplify both sides of the equation.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Simplify $2 \frac{\ln (2 \ln (1 + \sin^{2} (0)) + C)}{2}$ .

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Step 6.3.1.1.1

Cancel the common factor of $2$ .

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Step 6.3.1.1.1.1

Cancel the common factor.

$2 \frac{\ln (2 \ln (1 + \sin^{2} (0)) + C)}{2} = 2 \cdot 1$

Step 6.3.1.1.1.2

Rewrite the expression.

$\ln (2 \ln (1 + \sin^{2} (0)) + C) = 2 \cdot 1$

Step 6.3.1.1.2

Simplify each term.

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Step 6.3.1.1.2.1

Simplify each term.

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Step 6.3.1.1.2.1.1

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

$\ln (2 \ln (1 + 0^{2}) + C) = 2 \cdot 1$

Step 6.3.1.1.2.1.2

Raising $0$ to any positive power yields $0$ .

$\ln (2 \ln (1 + 0) + C) = 2 \cdot 1$

Step 6.3.1.1.2.2

Add $1$ and $0$ .

$\ln (2 \ln (1) + C) = 2 \cdot 1$

Step 6.3.1.1.2.3

The natural logarithm of $1$ is $0$ .

$\ln (2 \cdot 0 + C) = 2 \cdot 1$

Step 6.3.1.1.2.4

Multiply $2$ by $0$ .

$\ln (0 + C) = 2 \cdot 1$

Step 6.3.1.1.3

Add $0$ and $C$ .

$\ln (C) = 2 \cdot 1$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Multiply $2$ by $1$ .

$\ln (C) = 2$

Step 6.4

To solve for $C$ , rewrite the equation using properties of logarithms.

$e^{\ln (C)} = e^{2}$

Step 6.5

Rewrite $\ln (C) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{2} = C$

Step 6.6

Rewrite the equation as $C = e^{2}$ .

$C = e^{2}$

Step 7

Substitute $e^{2}$ for $C$ in $y = \frac{\ln (2 \ln (1 + \sin^{2} (x)) + C)}{2}$ and simplify.

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Step 7.1

Substitute $e^{2}$ for $C$ .

$y = \frac{\ln (2 \ln (1 + \sin^{2} (x)) + e^{2})}{2}$

Step 7.2

Rewrite $\frac{\ln (2 \ln (1 + \sin^{2} (x)) + e^{2})}{2}$ as $\frac{1}{2} \ln (2 \ln (1 + \sin^{2} (x)) + e^{2})$ .

$y = \frac{1}{2} \ln (2 \ln (1 + \sin^{2} (x)) + e^{2})$

Step 7.3

Simplify $\frac{1}{2} \ln (2 \ln (1 + \sin^{2} (x)) + e^{2})$ by moving $\frac{1}{2}$ inside the logarithm.

$y = \ln ({(2 \ln (1 + \sin^{2} (x)) + e^{2})}^{\frac{1}{2}})$

Step 7.4

Simplify $2 \ln (1 + \sin^{2} (x))$ by moving $2$ inside the logarithm.

$y = \ln ({(\ln ({(1 + \sin^{2} (x))}^{2}) + e^{2})}^{\frac{1}{2}})$