Calculus Examples

$\int_{2 \sqrt{2}}^{4} \frac{1}{t^{3} \sqrt{t^{2} - 4}} d t$

Step 1

Let $t = 2 \sec (t_{1})$ , where $- \frac{π}{2} \leq t_{1} \leq \frac{π}{2}$ . Then $d t = 2 \sec (t_{1}) \tan (t_{1}) d t_{1}$ . Note that since $- \frac{π}{2} \leq t_{1} \leq \frac{π}{2}$ , $2 \sec (t_{1}) \tan (t_{1})$ is positive.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{{(2 \sec (t_{1}))}^{2} - 4}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2

Simplify terms.

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Step 2.1

Simplify $\sqrt{{(2 \sec (t_{1}))}^{2} - 4}$ .

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

Apply the product rule to $2 \sec (t_{1})$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{2^{2} \sec^{2} (t_{1}) - 4}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.1.2

Raise $2$ to the power of $2$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{4 \sec^{2} (t_{1}) - 4}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.2

Factor $4$ out of $4 \sec^{2} (t_{1})$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{4 (\sec^{2} (t_{1})) - 4}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.3

Factor $4$ out of $- 4$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{4 \sec^{2} (t_{1}) + 4 \cdot - 1}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.4

Factor $4$ out of $4 \sec^{2} (t_{1}) + 4 \cdot - 1$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{4 (\sec^{2} (t_{1}) - 1)}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.5

Apply pythagorean identity.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{4 \tan^{2} (t_{1})}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.6

Rewrite $4 \tan^{2} (t_{1})$ as ${(2 \tan (t_{1}))}^{2}$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} \sqrt{{(2 \tan (t_{1}))}^{2}}} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.1.7

Pull terms out from under the radical, assuming positive real numbers.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3} (2 \tan (t_{1}))} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.2

Simplify terms.

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Step 2.2.1

Cancel the common factor of $2 \tan (t_{1})$ .

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Step 2.2.1.1

Factor $2 \tan (t_{1})$ out of ${(2 \sec (t_{1}))}^{3} (2 \tan (t_{1}))$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{2 \tan (t_{1}) ({(2 \sec (t_{1}))}^{3})} (2 \sec (t_{1}) \tan (t_{1})) d t_{1}$

Step 2.2.1.2

Factor $2 \tan (t_{1})$ out of $2 \sec (t_{1}) \tan (t_{1})$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{2 \tan (t_{1}) ({(2 \sec (t_{1}))}^{3})} (2 \tan (t_{1}) (\sec (t_{1}))) d t_{1}$

Step 2.2.1.3

Cancel the common factor.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{2 \tan (t_{1}) {(2 \sec (t_{1}))}^{3}} (2 \tan (t_{1}) \sec (t_{1})) d t_{1}$

Step 2.2.1.4

Rewrite the expression.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{{(2 \sec (t_{1}))}^{3}} \sec (t_{1}) d t_{1}$

Step 2.2.2

Combine $\frac{1}{{(2 \sec (t_{1}))}^{3}}$ and $\sec (t_{1})$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1})}{{(2 \sec (t_{1}))}^{3}} d t_{1}$

Step 2.2.3

Simplify.

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Step 2.2.3.1

Factor $2$ out of $2 \sec (t_{1})$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1})}{{(2 (\sec (t_{1})))}^{3}} d t_{1}$

Step 2.2.3.2

Apply the product rule to $2 (\sec (t_{1}))$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1})}{2^{3} \sec^{3} (t_{1})} d t_{1}$

Step 2.2.3.3

Raise $2$ to the power of $3$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1})}{8 \sec^{3} (t_{1})} d t_{1}$

Step 2.2.3.4

Cancel the common factor of $\sec (t_{1})$ and $\sec^{3} (t_{1})$ .

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Step 2.2.3.4.1

Multiply by $1$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1}) \cdot 1}{8 \sec^{3} (t_{1})} d t_{1}$

Step 2.2.3.4.2

Cancel the common factors.

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Step 2.2.3.4.2.1

Factor $\sec (t_{1})$ out of $8 \sec^{3} (t_{1})$ .

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1}) \cdot 1}{\sec (t_{1}) (8 \sec^{2} (t_{1}))} d t_{1}$

Step 2.2.3.4.2.2

Cancel the common factor.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{\sec (t_{1}) \cdot 1}{\sec (t_{1}) (8 \sec^{2} (t_{1}))} d t_{1}$

Step 2.2.3.4.2.3

Rewrite the expression.

$\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{8 \sec^{2} (t_{1})} d t_{1}$

Step 3

Since $\frac{1}{8}$ is constant with respect to $t_{1}$ , move $\frac{1}{8}$ out of the integral.

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{\sec^{2} (t_{1})} d t_{1}$

Step 4

Simplify.

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Step 4.1

Rewrite $1$ as $1^{2}$ .

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1^{2}}{\sec^{2} (t_{1})} d t_{1}$

Step 4.2

Rewrite $\frac{1^{2}}{\sec^{2} (t_{1})}$ as ${(\frac{1}{\sec (t_{1})})}^{2}$ .

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} {(\frac{1}{\sec (t_{1})})}^{2} d t_{1}$

Step 4.3

Rewrite $\sec (t_{1})$ in terms of sines and cosines.

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} {(\frac{1}{\frac{1}{\cos (t_{1})}})}^{2} d t_{1}$

Step 4.4

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (t_{1})}$ .

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} {(1 \cos (t_{1}))}^{2} d t_{1}$

Step 4.5

Multiply $\cos (t_{1})$ by $1$ .

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} \cos^{2} (t_{1}) d t_{1}$

Step 5

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

$\frac{1}{8} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1 + \cos (2 t_{1})}{2} d t_{1}$

Step 6

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

$\frac{1}{8} (\frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{3}} 1 + \cos (2 t_{1}) d t_{1})$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{2}$ by $\frac{1}{8}$ .

$\frac{1}{2 \cdot 8} \int_{\frac{π}{4}}^{\frac{π}{3}} 1 + \cos (2 t_{1}) d t_{1}$

Step 7.2

Multiply $2$ by $8$ .

$\frac{1}{16} \int_{\frac{π}{4}}^{\frac{π}{3}} 1 + \cos (2 t_{1}) d t_{1}$

Step 8

Split the single integral into multiple integrals.

$\frac{1}{16} (\int_{\frac{π}{4}}^{\frac{π}{3}} d t_{1} + \int_{\frac{π}{4}}^{\frac{π}{3}} \cos (2 t_{1}) d t_{1})$

Step 9

Apply the constant rule.

$\frac{1}{16} (t_{1}]_{\frac{π}{4}}^{\frac{π}{3}} + \int_{\frac{π}{4}}^{\frac{π}{3}} \cos (2 t_{1}) d t_{1})$

Step 10

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

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Step 10.1

Let $u = 2 t_{1}$ . Find $\frac{d u}{d t_{1}}$ .

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Step 10.1.1

Differentiate $2 t_{1}$ .

$\frac{d}{d t_{1}} [2 t_{1}]$

Step 10.1.2

Since $2$ is constant with respect to $t_{1}$ , the derivative of $2 t_{1}$ with respect to $t_{1}$ is $2 \frac{d}{d t_{1}} [t_{1}]$ .

$2 \frac{d}{d t_{1}} [t_{1}]$

Step 10.1.3

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

$2 \cdot 1$

Step 10.1.4

Multiply $2$ by $1$ .

$2$

Step 10.2

Substitute the lower limit in for $t_{1}$ in $u = 2 t_{1}$ .

$u_{lower} = 2 \frac{π}{4}$

Step 10.3

Cancel the common factor of $2$ .

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Step 10.3.1

Factor $2$ out of $4$ .

$u_{lower} = 2 \frac{π}{2 (2)}$

Step 10.3.2

Cancel the common factor.

$u_{lower} = 2 \frac{π}{2 \cdot 2}$

Step 10.3.3

Rewrite the expression.

$u_{lower} = \frac{π}{2}$

Step 10.4

Substitute the upper limit in for $t_{1}$ in $u = 2 t_{1}$ .

$u_{upper} = 2 \frac{π}{3}$

Step 10.5

Combine $2$ and $\frac{π}{3}$ .

$u_{upper} = \frac{2 π}{3}$

Step 10.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = \frac{π}{2}$

$u_{upper} = \frac{2 π}{3}$

Step 10.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\frac{1}{16} (t_{1}]_{\frac{π}{4}}^{\frac{π}{3}} + \int_{\frac{π}{2}}^{\frac{2 π}{3}} \cos (u) \frac{1}{2} d u)$

Step 11

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

$\frac{1}{16} (t_{1}]_{\frac{π}{4}}^{\frac{π}{3}} + \int_{\frac{π}{2}}^{\frac{2 π}{3}} \frac{\cos (u)}{2} d u)$

Step 12

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

$\frac{1}{16} (t_{1}]_{\frac{π}{4}}^{\frac{π}{3}} + \frac{1}{2} \int_{\frac{π}{2}}^{\frac{2 π}{3}} \cos (u) d u)$

Step 13

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$\frac{1}{16} (t_{1}]_{\frac{π}{4}}^{\frac{π}{3}} + \frac{1}{2} \sin (u)]_{\frac{π}{2}}^{\frac{2 π}{3}})$

Step 14

Substitute and simplify.

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Step 14.1

Evaluate $t_{1}$ at $\frac{π}{3}$ and at $\frac{π}{4}$ .

$\frac{1}{16} ((\frac{π}{3}) - \frac{π}{4} + \frac{1}{2} \sin (u)]_{\frac{π}{2}}^{\frac{2 π}{3}})$

Step 14.2

Evaluate $\sin (u)$ at $\frac{2 π}{3}$ and at $\frac{π}{2}$ .

$\frac{1}{16} ((\frac{π}{3}) - \frac{π}{4} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3

Simplify.

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Step 14.3.1

To write $\frac{π}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{1}{16} (\frac{π}{3} \cdot \frac{4}{4} - \frac{π}{4} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.2

To write $- \frac{π}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{16} (\frac{π}{3} \cdot \frac{4}{4} - \frac{π}{4} \cdot \frac{3}{3} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.3

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

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Step 14.3.3.1

Multiply $\frac{π}{3}$ by $\frac{4}{4}$ .

$\frac{1}{16} (\frac{π \cdot 4}{3 \cdot 4} - \frac{π}{4} \cdot \frac{3}{3} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.3.2

Multiply $3$ by $4$ .

$\frac{1}{16} (\frac{π \cdot 4}{12} - \frac{π}{4} \cdot \frac{3}{3} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.3.3

Multiply $\frac{π}{4}$ by $\frac{3}{3}$ .

$\frac{1}{16} (\frac{π \cdot 4}{12} - \frac{π \cdot 3}{4 \cdot 3} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.3.4

Multiply $4$ by $3$ .

$\frac{1}{16} (\frac{π \cdot 4}{12} - \frac{π \cdot 3}{12} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.4

Combine the numerators over the common denominator.

$\frac{1}{16} (\frac{π \cdot 4 - π \cdot 3}{12} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.5

Move $4$ to the left of $π$ .

$\frac{1}{16} (\frac{4 \cdot π - π \cdot 3}{12} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.6

Multiply $3$ by $- 1$ .

$\frac{1}{16} (\frac{4 π - 3 π}{12} + \frac{1}{2} ((\sin (\frac{2 π}{3})) - \sin (\frac{π}{2})))$

Step 14.3.7

Subtract $3 π$ from $4 π$ .

$\frac{1}{16} (\frac{π}{12} + \frac{1}{2} (\sin (\frac{2 π}{3}) - \sin (\frac{π}{2})))$

Step 15

Simplify.

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Step 15.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$\frac{1}{16} (\frac{π}{12} + \frac{1}{2} (\sin (\frac{2 π}{3}) - 1 \cdot 1))$

Step 15.2

Multiply $- 1$ by $1$ .

$\frac{1}{16} (\frac{π}{12} + \frac{1}{2} (\sin (\frac{2 π}{3}) - 1))$

Step 16

Simplify.

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Step 16.1

Simplify each term.

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Step 16.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$\frac{1}{16} (\frac{π}{12} + \frac{1}{2} (\sin (\frac{π}{3}) - 1))$

Step 16.1.2

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{1}{16} (\frac{π}{12} + \frac{1}{2} (\frac{\sqrt{3}}{2} - 1))$

Step 16.2

Apply the distributive property.

$\frac{1}{16} (\frac{π}{12} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot - 1)$

Step 16.3

Multiply $\frac{1}{2} \cdot \frac{\sqrt{3}}{2}$ .

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Step 16.3.1

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

$\frac{1}{16} (\frac{π}{12} + \frac{\sqrt{3}}{2 \cdot 2} + \frac{1}{2} \cdot - 1)$

Step 16.3.2

Multiply $2$ by $2$ .

$\frac{1}{16} (\frac{π}{12} + \frac{\sqrt{3}}{4} + \frac{1}{2} \cdot - 1)$

Step 16.4

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

$\frac{1}{16} (\frac{π}{12} + \frac{\sqrt{3}}{4} + \frac{- 1}{2})$

Step 16.5

Move the negative in front of the fraction.

$\frac{1}{16} (\frac{π}{12} + \frac{\sqrt{3}}{4} - \frac{1}{2})$

Step 16.6

Apply the distributive property.

$\frac{1}{16} \cdot \frac{π}{12} + \frac{1}{16} \cdot \frac{\sqrt{3}}{4} + \frac{1}{16} (- \frac{1}{2})$

Step 16.7

Simplify.

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Step 16.7.1

Multiply $\frac{1}{16} \cdot \frac{π}{12}$ .

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Step 16.7.1.1

Multiply $\frac{1}{16}$ by $\frac{π}{12}$ .

$\frac{π}{16 \cdot 12} + \frac{1}{16} \cdot \frac{\sqrt{3}}{4} + \frac{1}{16} (- \frac{1}{2})$

Step 16.7.1.2

Multiply $16$ by $12$ .

$\frac{π}{192} + \frac{1}{16} \cdot \frac{\sqrt{3}}{4} + \frac{1}{16} (- \frac{1}{2})$

Step 16.7.2

Multiply $\frac{1}{16} \cdot \frac{\sqrt{3}}{4}$ .

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Step 16.7.2.1

Multiply $\frac{1}{16}$ by $\frac{\sqrt{3}}{4}$ .

$\frac{π}{192} + \frac{\sqrt{3}}{16 \cdot 4} + \frac{1}{16} (- \frac{1}{2})$

Step 16.7.2.2

Multiply $16$ by $4$ .

$\frac{π}{192} + \frac{\sqrt{3}}{64} + \frac{1}{16} (- \frac{1}{2})$

Step 16.7.3

Multiply $\frac{1}{16} (- \frac{1}{2})$ .

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Step 16.7.3.1

Multiply $\frac{1}{16}$ by $\frac{1}{2}$ .

$\frac{π}{192} + \frac{\sqrt{3}}{64} - \frac{1}{16 \cdot 2}$

Step 16.7.3.2

Multiply $16$ by $2$ .

$\frac{π}{192} + \frac{\sqrt{3}}{64} - \frac{1}{32}$

Step 17

The result can be shown in multiple forms.

Exact Form:

$\frac{π}{192} + \frac{\sqrt{3}}{64} - \frac{1}{32}$

Decimal Form:

$0.01217575 \dots$

Step 18