Calculus Examples

$\frac{2}{\sqrt{x + 3}} - \sin^{2} (2 x)$

Step 1

Write $\frac{2}{\sqrt{x + 3}} - \sin^{2} (2 x)$ as a function.

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

Step 2

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 3

Set up the integral to solve.

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

Step 4

Split the single integral into multiple integrals.

$\int \frac{2}{\sqrt{x + 3}} d x + \int - \sin^{2} (2 x) d x$

Step 5

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

$2 \int \frac{1}{\sqrt{x + 3}} d x + \int - \sin^{2} (2 x) d x$

Step 6

Let $u_{1} = x + 3$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 6.1

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

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Step 6.1.1

Differentiate $x + 3$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

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

Step 6.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

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

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

Step 7

Apply basic rules of exponents.

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

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

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Step 7.3.1

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

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

Step 7.3.2

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

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

Step 7.3.3

Move the negative in front of the fraction.

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

Step 8

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

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

Step 9

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 10

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

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Step 10.1

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

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Step 10.1.1

Differentiate $2 x$ .

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

Step 10.1.2

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

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

Step 10.1.3

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

$2 \cdot 1$

Step 10.1.4

Multiply $2$ by $1$ .

$2$

Step 10.2

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \int \sin^{2} (u_{2}) \frac{1}{2} d u_{2}$

Step 11

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \int \frac{\sin^{2} (u_{2})}{2} d u_{2}$

Step 12

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - (\frac{1}{2} \int \sin^{2} (u_{2}) d u_{2})$

Step 13

Use the half-angle formula to rewrite $\sin^{2} (u_{2})$ as $\frac{1 - \cos (2 u_{2})}{2}$ .

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{2} \int \frac{1 - \cos (2 u_{2})}{2} d u_{2}$

Step 14

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{2} (\frac{1}{2} \int 1 - \cos (2 u_{2}) d u_{2})$

Step 15

Simplify.

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Step 15.1

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{2 \cdot 2} \int 1 - \cos (2 u_{2}) d u_{2}$

Step 15.2

Multiply $2$ by $2$ .

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} \int 1 - \cos (2 u_{2}) d u_{2}$

Step 16

Split the single integral into multiple integrals.

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} (\int d u_{2} + \int - \cos (2 u_{2}) d u_{2})$

Step 17

Apply the constant rule.

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} (u_{2} + C + \int - \cos (2 u_{2}) d u_{2})$

Step 18

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} (u_{2} + C - \int \cos (2 u_{2}) d u_{2})$

Step 19

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

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Step 19.1

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

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Step 19.1.1

Differentiate $2 u_{2}$ .

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

Step 19.1.2

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

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

Step 19.1.3

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

$2 \cdot 1$

Step 19.1.4

Multiply $2$ by $1$ .

$2$

Step 19.2

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

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} (u_{2} + C - \int \cos (u_{3}) \frac{1}{2} d u_{3})$

Step 20

Combine $\cos (u_{3})$ and $\frac{1}{2}$ .

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} (u_{2} + C - \int \frac{\cos (u_{3})}{2} d u_{3})$

Step 21

Since $\frac{1}{2}$ is constant with respect to $u_{3}$ , move $\frac{1}{2}$ out of the integral.

$2 (2 {u_{1}}^{\frac{1}{2}} + C) - \frac{1}{4} (u_{2} + C - (\frac{1}{2} \int \cos (u_{3}) d u_{3}))$

Step 22

The integral of $\cos (u_{3})$ with respect to $u_{3}$ is $\sin (u_{3})$ .

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

Step 23

Simplify.

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

Step 24

Substitute back in for each integration substitution variable.

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Step 24.1

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

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

Step 24.2

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

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

Step 24.3

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

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

Step 24.4

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

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

Step 25

Simplify.

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Step 25.1

Simplify each term.

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Step 25.1.1

Multiply $2$ by $2$ .

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

Step 25.1.2

Combine $\sin (4 x)$ and $\frac{1}{2}$ .

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

Step 25.2

Apply the distributive property.

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

Step 25.3

Cancel the common factor of $2$ .

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Step 25.3.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

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

Step 25.3.2

Factor $2$ out of $4$ .

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

Step 25.3.3

Factor $2$ out of $2 x$ .

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

Step 25.3.4

Cancel the common factor.

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

Step 25.3.5

Rewrite the expression.

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

Step 25.4

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

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

Step 25.5

Multiply $- \frac{1}{4} (- \frac{\sin (4 x)}{2})$ .

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Step 25.5.1

Multiply $- 1$ by $- 1$ .

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

Step 25.5.2

Multiply $\frac{1}{4}$ by $1$ .

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

Step 25.5.3

Multiply $\frac{1}{4}$ by $\frac{\sin (4 x)}{2}$ .

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

Step 25.5.4

Multiply $4$ by $2$ .

$4 {(x + 3)}^{\frac{1}{2}} + \frac{- x}{2} + \frac{\sin (4 x)}{8} + C$

Step 25.6

Move the negative in front of the fraction.

$4 {(x + 3)}^{\frac{1}{2}} - \frac{x}{2} + \frac{\sin (4 x)}{8} + C$

Step 26

Reorder terms.

$4 {(x + 3)}^{\frac{1}{2}} - \frac{1}{2} x + \frac{1}{8} \sin (4 x) + C$

Step 27

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

$F (x) =$ $4 {(x + 3)}^{\frac{1}{2}} - \frac{1}{2} x + \frac{1}{8} \sin (4 x) + C$