Calculus Examples

$f (x) = \frac{5}{2 \sqrt{3 x + 2}} + \frac{1}{\sin^{2} (4 x)}$

Step 1

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 2

Set up the integral to solve.

$F (x) = \int \frac{5}{2 \sqrt{3 x + 2}} + \frac{1}{\sin^{2} (4 x)} d x$

Step 3

Convert from $\frac{1}{\sin^{2} (4 x)}$ to $\csc^{2} (4 x)$ .

$\int \frac{5}{2 \sqrt{3 x + 2}} + \csc^{2} (4 x) d x$

Step 4

Split the single integral into multiple integrals.

$\int \frac{5}{2 \sqrt{3 x + 2}} d x + \int \csc^{2} (4 x) d x$

Step 5

Since $\frac{5}{2}$ is constant with respect to $x$ , move $\frac{5}{2}$ out of the integral.

$\frac{5}{2} \int \frac{1}{\sqrt{3 x + 2}} d x + \int \csc^{2} (4 x) d x$

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $3 x + 2$ .

$\frac{d}{d x} [3 x + 2]$

Step 6.1.2

By the Sum Rule, the derivative of $3 x + 2$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$

Step 6.1.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 6.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 6.1.3.2

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

$3 \cdot 1 + \frac{d}{d x} [2]$

Step 6.1.3.3

Multiply $3$ by $1$ .

$3 + \frac{d}{d x} [2]$

Step 6.1.4

Differentiate using the Constant Rule.

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Step 6.1.4.1

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

$3 + 0$

Step 6.1.4.2

Add $3$ and $0$ .

$3$

Step 6.2

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

$\frac{5}{2} \int \frac{1}{\sqrt{u_{1}}} \cdot \frac{1}{3} d u_{1} + \int \csc^{2} (4 x) d x$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{\sqrt{u_{1}}}$ by $\frac{1}{3}$ .

$\frac{5}{2} \int \frac{1}{\sqrt{u_{1}} \cdot 3} d u_{1} + \int \csc^{2} (4 x) d x$

Step 7.2

Move $3$ to the left of $\sqrt{u_{1}}$ .

$\frac{5}{2} \int \frac{1}{3 \sqrt{u_{1}}} d u_{1} + \int \csc^{2} (4 x) d x$

Step 8

Since $\frac{1}{3}$ is constant with respect to $u_{1}$ , move $\frac{1}{3}$ out of the integral.

$\frac{5}{2} (\frac{1}{3} \int \frac{1}{\sqrt{u_{1}}} d u_{1}) + \int \csc^{2} (4 x) d x$

Step 9

Simplify the expression.

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Step 9.1

Simplify.

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Step 9.1.1

Multiply $\frac{1}{3}$ by $\frac{5}{2}$ .

$\frac{5}{3 \cdot 2} \int \frac{1}{\sqrt{u_{1}}} d u_{1} + \int \csc^{2} (4 x) d x$

Step 9.1.2

Multiply $3$ by $2$ .

$\frac{5}{6} \int \frac{1}{\sqrt{u_{1}}} d u_{1} + \int \csc^{2} (4 x) d x$

Step 9.2

Apply basic rules of exponents.

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Step 9.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u_{1}}$ as ${u_{1}}^{\frac{1}{2}}$ .

$\frac{5}{6} \int \frac{1}{{u_{1}}^{\frac{1}{2}}} d u_{1} + \int \csc^{2} (4 x) d x$

Step 9.2.2

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

$\frac{5}{6} \int {({u_{1}}^{\frac{1}{2}})}^{- 1} d u_{1} + \int \csc^{2} (4 x) d x$

Step 9.2.3

Multiply the exponents in ${({u_{1}}^{\frac{1}{2}})}^{- 1}$ .

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Step 9.2.3.1

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

$\frac{5}{6} \int {u_{1}}^{\frac{1}{2} \cdot - 1} d u_{1} + \int \csc^{2} (4 x) d x$

Step 9.2.3.2

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

$\frac{5}{6} \int {u_{1}}^{\frac{- 1}{2}} d u_{1} + \int \csc^{2} (4 x) d x$

Step 9.2.3.3

Move the negative in front of the fraction.

$\frac{5}{6} \int {u_{1}}^{- \frac{1}{2}} d u_{1} + \int \csc^{2} (4 x) d x$

Step 10

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

$\frac{5}{6} (2 {u_{1}}^{\frac{1}{2}} + C) + \int \csc^{2} (4 x) d x$

Step 11

Let $u_{2} = 4 x$ . Then $d u_{2} = 4 d x$ , so $\frac{1}{4} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 11.1

Let $u_{2} = 4 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 11.1.1

Differentiate $4 x$ .

$\frac{d}{d x} [4 x]$

Step 11.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x]$

Step 11.1.3

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

$4 \cdot 1$

Step 11.1.4

Multiply $4$ by $1$ .

$4$

Step 11.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{5}{6} (2 {u_{1}}^{\frac{1}{2}} + C) + \int \csc^{2} (u_{2}) \frac{1}{4} d u_{2}$

Step 12

Combine $\csc^{2} (u_{2})$ and $\frac{1}{4}$ .

$\frac{5}{6} (2 {u_{1}}^{\frac{1}{2}} + C) + \int \frac{\csc^{2} (u_{2})}{4} d u_{2}$

Step 13

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

$\frac{5}{6} (2 {u_{1}}^{\frac{1}{2}} + C) + \frac{1}{4} \int \csc^{2} (u_{2}) d u_{2}$

Step 14

Since the derivative of $- \cot (u_{2})$ is $\csc^{2} (u_{2})$ , the integral of $\csc^{2} (u_{2})$ is $- \cot (u_{2})$ .

$\frac{5}{6} (2 {u_{1}}^{\frac{1}{2}} + C) + \frac{1}{4} (- \cot (u_{2}) + C)$

Step 15

Simplify.

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Step 15.1

Simplify.

$\frac{5 {u_{1}}^{\frac{1}{2}}}{3} + \frac{1}{4} (- \cot (u_{2})) + C$

Step 15.2

Combine $\frac{1}{4}$ and $\cot (u_{2})$ .

$\frac{5 {u_{1}}^{\frac{1}{2}}}{3} - \frac{\cot (u_{2})}{4} + C$

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

Replace all occurrences of $u_{1}$ with $3 x + 2$ .

$\frac{5 {(3 x + 2)}^{\frac{1}{2}}}{3} - \frac{\cot (u_{2})}{4} + C$

Step 16.2

Replace all occurrences of $u_{2}$ with $4 x$ .

$\frac{5 {(3 x + 2)}^{\frac{1}{2}}}{3} - \frac{\cot (4 x)}{4} + C$

Step 17

Reorder terms.

$\frac{5}{3} {(3 x + 2)}^{\frac{1}{2}} - \frac{1}{4} \cot (4 x) + C$

Step 18

The answer is the antiderivative of the function $f (x) = \frac{5}{2 \sqrt{3 x + 2}} + \frac{1}{\sin^{2} (4 x)}$ .

$F (x) =$ $\frac{5}{3} {(3 x + 2)}^{\frac{1}{2}} - \frac{1}{4} \cot (4 x) + C$