Calculus Examples

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

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

Step 1

Let

u = 1 + cos (x)

$u = 1 + \cos (x)$ . Then

d u = - sin (x) d x

$d u = - \sin (x) d x$ , so

- d u = sin (x) d x

$- d u = \sin (x) d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 1.1

Let

u = 1 + cos (x)

$u = 1 + \cos (x)$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate

1 + cos (x)

$1 + \cos (x)$ .

\frac{d}{d x} [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)

$1 + \cos (x)$ with respect to

x

$x$ is

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

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

\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

$1$ is constant with respect to

x

$x$ , the derivative of

1

$1$ with respect to

x

$x$ is

0

$0$ .

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

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

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

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

Step 1.1.3

The derivative of

cos (x)

$\cos (x)$ with respect to

x

$x$ is

- sin (x)

$- \sin (x)$ .

0 - sin (x)

$0 - \sin (x)$

Step 1.1.4

Subtract

sin (x)

$\sin (x)$ from

0

$0$ .

- sin (x)

$- \sin (x)$

- sin (x)

$- \sin (x)$

Step 1.2

Substitute the lower limit in for

x

$x$ in

u = 1 + cos (x)

$u = 1 + \cos (x)$ .

u_{lower} = 1 + cos (0)

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

Step 1.3

Simplify.

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Step 1.3.1

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

u_{lower} = 1 + 1

$u_{lower} = 1 + 1$

Step 1.3.2

Add

1

$1$ and

1

$1$ .

u_{lower} = 2

$u_{lower} = 2$

u_{lower} = 2

$u_{lower} = 2$

Step 1.4

Substitute the upper limit in for

x

$x$ in

u = 1 + cos (x)

$u = 1 + \cos (x)$ .

u_{upper} = 1 + cos (π)

$u_{upper} = 1 + \cos (π)$

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)

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

Step 1.5.1.2

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

u_{upper} = 1 - 1 \cdot 1

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

Step 1.5.1.3

Multiply

- 1

$- 1$ by

1

$1$ .

u_{upper} = 1 - 1

$u_{upper} = 1 - 1$

u_{upper} = 1 - 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