Calculus Examples

\int \cos (2 y) d y

$\int \cos (2 y) d y$

Step 1

Let

u = 2 y

$u = 2 y$ . Then

d u = 2 d y

$d u = 2 d y$ , so

\frac{1}{2} d u = d y

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

u

$u$ and

d

$d$

u

$u$ .

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Step 1.1

Let

u = 2 y

$u = 2 y$ . Find

\frac{d u}{d y}

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

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Step 1.1.1

Differentiate

2 y

$2 y$ .

\frac{d}{d y} [2 y]

$\frac{d}{d y} [2 y]$

Step 1.1.2

Since

2

$2$ is constant with respect to

y

$y$ , the derivative of

2 y

$2 y$ with respect to

y

$y$ is

2 \frac{d}{d y} [y]

$2 \frac{d}{d y} [y]$ .

2 \frac{d}{d y} [y]

$2 \frac{d}{d y} [y]$

Step 1.1.3

Differentiate using the Power Rule which states that

\frac{d}{d y} [y^{n}]

$\frac{d}{d y} [y^{n}]$ is

n y^{n - 1}

$n y^{n - 1}$ where

n = 1

$n = 1$ .

2 \cdot 1

$2 \cdot 1$

Step 1.1.4

Multiply

2

$2$ by

1

$1$ .

2

$2$

2

$2$

Step 1.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

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

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

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

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

Step 2

Combine

\cos (u)

$\cos (u)$ and

\frac{1}{2}

$\frac{1}{2}$ .

\int \frac{\cos (u)}{2} d u

$\int \frac{\cos (u)}{2} d u$

Step 3

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} \int \cos (u) d u

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

Step 4

The integral of

\cos (u)

$\cos (u)$ with respect to

u

$u$ is

\sin (u)

$\sin (u)$ .

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

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

Step 5

Simplify.

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

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

Step 6

Replace all occurrences of

u

$u$ with

2 y

$2 y$ .

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

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

\int_{}^{} \cos (2 y) d y

$\int_{}^{} \cos (2 y) d y$

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$