Calculus Examples

\int {cos}^{2} (θ) d θ

$\int \cos^{2} (θ) d θ$

Step 1

Use the half-angle formula to rewrite

{cos}^{2} (θ)

$\cos^{2} (θ)$ as

\frac{1 + cos (2 θ)}{2}

$\frac{1 + \cos (2 θ)}{2}$ .

\int \frac{1 + cos (2 θ)}{2} d θ

$\int \frac{1 + \cos (2 θ)}{2} d θ$

Step 2

Since

\frac{1}{2}

$\frac{1}{2}$ is constant with respect to

θ

$θ$ , move

\frac{1}{2}

$\frac{1}{2}$ out of the integral.

\frac{1}{2} \int 1 + cos (2 θ) d θ

$\frac{1}{2} \int 1 + \cos (2 θ) d θ$

Step 3

Split the single integral into multiple integrals.

\frac{1}{2} (\int d θ + \int cos (2 θ) d θ)

$\frac{1}{2} (\int d θ + \int \cos (2 θ) d θ)$

Step 4

Apply the constant rule.

\frac{1}{2} (θ + C + \int cos (2 θ) d θ)

$\frac{1}{2} (θ + C + \int \cos (2 θ) d θ)$

Step 5

Let

u = 2 θ

$u = 2 θ$ . Then

d u = 2 d θ

$d u = 2 d θ$ , so

\frac{1}{2} d u = d θ

$\frac{1}{2} d u = d θ$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 5.1

Let

u = 2 θ

$u = 2 θ$ . Find

\frac{d u}{d θ}

$\frac{d u}{d θ}$ .

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Step 5.1.1

Differentiate

2 θ

$2 θ$ .

\frac{d}{d θ} [2 θ]

$\frac{d}{d θ} [2 θ]$

Step 5.1.2

Since

2

$2$ is constant with respect to

θ

$θ$ , the derivative of

2 θ

$2 θ$ with respect to

θ

$θ$ is

2 \frac{d}{d θ} [θ]

$2 \frac{d}{d θ} [θ]$ .

2 \frac{d}{d θ} [θ]

$2 \frac{d}{d θ} [θ]$

Step 5.1.3

Differentiate using the Power Rule which states that

\frac{d}{d θ} [θ^{n}]

$\frac{d}{d θ} [θ^{n}]$ is

n θ^{n - 1}

$n θ^{n - 1}$ where

n = 1

$n = 1$ .

2 \cdot 1

$2 \cdot 1$

Step 5.1.4

Multiply

2

$2$ by

1

$1$ .

2

$2$

2

$2$

Step 5.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

\frac{1}{2} (θ + C + \int cos (u) \frac{1}{2} d u)

$\frac{1}{2} (θ + C + \int \cos (u) \frac{1}{2} d u)$

\frac{1}{2} (θ + C + \int cos (u) \frac{1}{2} d u)

$\frac{1}{2} (θ + C + \int \cos (u) \frac{1}{2} d u)$

Step 6

Combine

cos (u)

$\cos (u)$ and

\frac{1}{2}

$\frac{1}{2}$ .

\frac{1}{2} (θ + C + \int \frac{cos (u)}{2} d u)

$\frac{1}{2} (θ + C + \int \frac{\cos (u)}{2} d u)$

Step 7

Since

\frac{1}{2}

$\frac{1}{2}$ is constant with respect to

u

$u$ , move

\frac{1}{2}

$\frac{1}{2}$ out of the integral.

\frac{1}{2} (θ + C + \frac{1}{2} \int cos (u) d u)

$\frac{1}{2} (θ + C + \frac{1}{2} \int \cos (u) d u)$

Step 8

The integral of

cos (u)

$\cos (u)$ with respect to

u

$u$ is

sin (u)

$\sin (u)$ .

\frac{1}{2} (θ + C + \frac{1}{2} (sin (u) + C))

$\frac{1}{2} (θ + C + \frac{1}{2} (\sin (u) + C))$

Step 9

Simplify.

\frac{1}{2} (θ + \frac{1}{2} sin (u)) + C

$\frac{1}{2} (θ + \frac{1}{2} \sin (u)) + C$

Step 10

Replace all occurrences of

u

$u$ with

2 θ

$2 θ$ .

\frac{1}{2} (θ + \frac{1}{2} sin (2 θ)) + C

$\frac{1}{2} (θ + \frac{1}{2} \sin (2 θ)) + C$

Step 11

Simplify.

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Step 11.1

Combine

\frac{1}{2}

$\frac{1}{2}$ and

sin (2 θ)

$\sin (2 θ)$ .

\frac{1}{2} (θ + \frac{sin (2 θ)}{2}) + C

$\frac{1}{2} (θ + \frac{\sin (2 θ)}{2}) + C$

Step 11.2

Apply the distributive property.

\frac{1}{2} θ + \frac{1}{2} \cdot \frac{sin (2 θ)}{2} + C

$\frac{1}{2} θ + \frac{1}{2} \cdot \frac{\sin (2 θ)}{2} + C$

Step 11.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

θ

$θ$ .

\frac{θ}{2} + \frac{1}{2} \cdot \frac{sin (2 θ)}{2} + C

$\frac{θ}{2} + \frac{1}{2} \cdot \frac{\sin (2 θ)}{2} + C$

Step 11.4

Multiply

\frac{1}{2} \cdot \frac{sin (2 θ)}{2}

$\frac{1}{2} \cdot \frac{\sin (2 θ)}{2}$ .

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Step 11.4.1

Multiply

\frac{1}{2}

$\frac{1}{2}$ by

\frac{sin (2 θ)}{2}

$\frac{\sin (2 θ)}{2}$ .

\frac{θ}{2} + \frac{sin (2 θ)}{2 \cdot 2} + C

$\frac{θ}{2} + \frac{\sin (2 θ)}{2 \cdot 2} + C$

Step 11.4.2

Multiply

2

$2$ by

2

$2$ .

\frac{θ}{2} + \frac{sin (2 θ)}{4} + C

$\frac{θ}{2} + \frac{\sin (2 θ)}{4} + C$

\frac{θ}{2} + \frac{sin (2 θ)}{4} + C

$\frac{θ}{2} + \frac{\sin (2 θ)}{4} + C$

\frac{θ}{2} + \frac{sin (2 θ)}{4} + C

$\frac{θ}{2} + \frac{\sin (2 θ)}{4} + C$

Step 12

Reorder terms.

\frac{1}{2} θ + \frac{1}{4} sin (2 θ) + C

$\frac{1}{2} θ + \frac{1}{4} \sin (2 θ) + C$