Calculus Examples

$\int_{0}^{\frac{π}{2}} \frac{\cos (t)}{\sqrt{1 + \sin^{2} (t)}} d t$

Step 1

Let $u = \sin (t)$ . Then $d u = \cos (t) d t$ , so $\frac{1}{\cos (t)} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = \sin (t)$ . Find $\frac{d u}{d t}$ .

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Step 1.1.1

Differentiate $\sin (t)$ .

$\frac{d}{d t} [\sin (t)]$

Step 1.1.2

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

$\cos (t)$

Step 1.2

Substitute the lower limit in for $t$ in $u = \sin (t)$ .

$u_{lower} = \sin (0)$

Step 1.3

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

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $t$ in $u = \sin (t)$ .

$u_{upper} = \sin (\frac{π}{2})$

Step 1.5

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

$u_{upper} = 1$

Step 1.6

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

$u_{lower} = 0$

$u_{upper} = 1$

Step 1.7

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

$\int_{0}^{1} \frac{1}{\sqrt{1 + u^{2}}} d u$

Step 2

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

$\int_{0}^{\frac{π}{4}} \frac{1}{\sqrt{1 + \tan^{2} (t_{1})}} \sec^{2} (t_{1}) d t_{1}$

Step 3

Simplify terms.

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Step 3.1

Simplify $\sqrt{1 + \tan^{2} (t_{1})}$ .

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Step 3.1.1

Rearrange terms.

$\int_{0}^{\frac{π}{4}} \frac{1}{\sqrt{\tan^{2} (t_{1}) + 1}} \sec^{2} (t_{1}) d t_{1}$

Step 3.1.2

Apply pythagorean identity.

$\int_{0}^{\frac{π}{4}} \frac{1}{\sqrt{\sec^{2} (t_{1})}} \sec^{2} (t_{1}) d t_{1}$

Step 3.1.3

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

$\int_{0}^{\frac{π}{4}} \frac{1}{\sec (t_{1})} \sec^{2} (t_{1}) d t_{1}$

Step 3.2

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

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Step 3.2.1

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

$\int_{0}^{\frac{π}{4}} \frac{1}{\sec (t_{1})} (\sec (t_{1}) \sec (t_{1})) d t_{1}$

Step 3.2.2

Cancel the common factor.

$\int_{0}^{\frac{π}{4}} \frac{1}{\sec (t_{1})} (\sec (t_{1}) \sec (t_{1})) d t_{1}$

Step 3.2.3

Rewrite the expression.

$\int_{0}^{\frac{π}{4}} \sec (t_{1}) d t_{1}$

Step 4

The integral of $\sec (t_{1})$ with respect to $t_{1}$ is $\ln (| \sec (t_{1}) + \tan (t_{1}) |)$ .

$\ln (| \sec (t_{1}) + \tan (t_{1}) |)]_{0}^{\frac{π}{4}}$

Step 5

Evaluate $\ln (| \sec (t_{1}) + \tan (t_{1}) |)$ at $\frac{π}{4}$ and at $0$ .

$\ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |)$

Step 6

Simplify.

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Step 6.1

The exact value of $\sec (\frac{π}{4})$ is $\frac{2}{\sqrt{2}}$ .

$\ln (| \frac{2}{\sqrt{2}} + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |)$

Step 6.2

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$\ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| \sec (0) + \tan (0) |)$

Step 6.3

The exact value of $\sec (0)$ is $1$ .

$\ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + \tan (0) |)$

Step 6.4

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

$\ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |)$

Step 6.5

Add $1$ and $0$ .

$\ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 |)$

Step 6.6

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| \frac{2}{\sqrt{2}} + 1 |}{| 1 |})$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Simplify each term.

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Step 7.1.1.1

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

$\ln (\frac{| \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} + 1 |}{| 1 |})$

Step 7.1.1.2

Combine and simplify the denominator.

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Step 7.1.1.2.1

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

$\ln (\frac{| \frac{2 \sqrt{2}}{\sqrt{2} \sqrt{2}} + 1 |}{| 1 |})$

Step 7.1.1.2.2

Raise $\sqrt{2}$ to the power of $1$ .

$\ln (\frac{| \frac{2 \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}} + 1 |}{| 1 |})$

Step 7.1.1.2.3

Raise $\sqrt{2}$ to the power of $1$ .

$\ln (\frac{| \frac{2 \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}} + 1 |}{| 1 |})$

Step 7.1.1.2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\ln (\frac{| \frac{2 \sqrt{2}}{{\sqrt{2}}^{1 + 1}} + 1 |}{| 1 |})$

Step 7.1.1.2.5

Add $1$ and $1$ .

$\ln (\frac{| \frac{2 \sqrt{2}}{{\sqrt{2}}^{2}} + 1 |}{| 1 |})$

Step 7.1.1.2.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 7.1.1.2.6.1

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

$\ln (\frac{| \frac{2 \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}} + 1 |}{| 1 |})$

Step 7.1.1.2.6.2

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

$\ln (\frac{| \frac{2 \sqrt{2}}{2^{\frac{1}{2} \cdot 2}} + 1 |}{| 1 |})$

Step 7.1.1.2.6.3

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

$\ln (\frac{| \frac{2 \sqrt{2}}{2^{\frac{2}{2}}} + 1 |}{| 1 |})$

Step 7.1.1.2.6.4

Cancel the common factor of $2$ .

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Step 7.1.1.2.6.4.1

Cancel the common factor.

$\ln (\frac{| \frac{2 \sqrt{2}}{2^{\frac{2}{2}}} + 1 |}{| 1 |})$

Step 7.1.1.2.6.4.2

Rewrite the expression.

$\ln (\frac{| \frac{2 \sqrt{2}}{2^{1}} + 1 |}{| 1 |})$

Step 7.1.1.2.6.5

Evaluate the exponent.

$\ln (\frac{| \frac{2 \sqrt{2}}{2} + 1 |}{| 1 |})$

Step 7.1.1.3

Cancel the common factor of $2$ .

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Step 7.1.1.3.1

Cancel the common factor.

$\ln (\frac{| \frac{2 \sqrt{2}}{2} + 1 |}{| 1 |})$

Step 7.1.1.3.2

Divide $\sqrt{2}$ by $1$ .

$\ln (\frac{| \sqrt{2} + 1 |}{| 1 |})$

Step 7.1.2

$\sqrt{2} + 1$ is approximately $2.41421356$ which is positive so remove the absolute value

$\ln (\frac{\sqrt{2} + 1}{| 1 |})$

Step 7.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\ln (\frac{\sqrt{2} + 1}{1})$

Step 7.3

Divide $\sqrt{2} + 1$ by $1$ .

$\ln (\sqrt{2} + 1)$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\ln (\sqrt{2} + 1)$

Decimal Form:

$0.88137358 \dots$