Calculus Examples

$\cos (x) - \sin (2 x)$

Write $\cos (x) - \sin (2 x)$ as a function.

$f (x) = \cos (x) - \sin (2 x)$

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Set up the integral to solve.

$F (x) = \int \cos (x) - \sin (2 x) d x$

Split the single integral into multiple integrals.

$\int \cos (x) d x + \int - \sin (2 x) d x$

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

$\sin (x) + C + \int - \sin (2 x) d x$

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

$\sin (x) + C - \int \sin (2 x) d x$

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = 2 x$ . Find $\frac{d u}{d x}$ .

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Differentiate $2 x$ .

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

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1$

Multiply $2$ by $1$ .

$2$

Rewrite the problem using $u$ and $d u$ .

$\sin (x) + C - \int \sin (u) \frac{1}{2} d u$

Combine $\sin (u)$ and $\frac{1}{2}$ .

$\sin (x) + C - \int \frac{\sin (u)}{2} d u$

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\sin (x) + C - (\frac{1}{2} \int \sin (u) d u)$

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

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

Simplify.

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

Replace all occurrences of $u$ with $2 x$ .

$\sin (x) + \frac{\cos (2 x)}{2} + C$

Reorder terms.

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

The answer is the antiderivative of the function $f (x) = \cos (x) - \sin (2 x)$ .

$F (x) =$ $\sin (x) + \frac{1}{2} \cos (2 x) + C$