Calculus Examples

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

Step 1

Let $u = 1 + \cos (x)$ . Then $d u = - \sin (x) d x$ , so $- d u = \sin (x) d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 1 + \cos (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $1 + \cos (x)$ .

$\frac{d}{d x} [1 + \cos (x)]$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 + \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\cos (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\cos (x)]$

Step 1.1.2.2

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

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

Step 1.1.3

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

$0 - \sin (x)$

Step 1.1.4

Subtract $\sin (x)$ from $0$ .

$- \sin (x)$

Step 1.2

Substitute the lower limit in for $x$ in $u = 1 + \cos (x)$ .

$u_{lower} = 1 + \cos (0)$

Step 1.3

Simplify.

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Step 1.3.1

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

$u_{lower} = 1 + 1$

Step 1.3.2

Add $1$ and $1$ .

$u_{lower} = 2$

Step 1.4

Substitute the upper limit in for $x$ in $u = 1 + \cos (x)$ .

$u_{upper} = 1 + \cos (π)$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$u_{upper} = 1 - \cos (0)$

Step 1.5.1.2

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

$u_{upper} = 1 - 1 \cdot 1$

Step 1.5.1.3

Multiply $- 1$ by $1$ .

$u_{upper} = 1 - 1$

Step 1.5.2

Subtract $1$ from $1$ .

$u_{upper} = 0$

Step 1.6

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

$u_{lower} = 2$

$u_{upper} = 0$

Step 1.7

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

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

Step 2

Split the fraction into multiple fractions.

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

Step 3

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

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

Step 4

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

$- (\ln (| u |)]_{2}^{0})$

Step 5

Evaluate $\ln (| u |)$ at $0$ and at $2$ .

$- (\ln (| 0 |) - \ln (| 2 |))$

Step 6

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

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

Step 7

Simplify.

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Step 7.1

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

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

Step 7.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$- \ln (\frac{0}{2})$

Step 7.3

Divide $0$ by $2$ .

$- \ln (0)$

Step 7.4

The natural logarithm of zero is undefined.

Undefined

$- \ln (0)$

Step 8

The natural logarithm of zero is undefined.

Undefined