Calculus Examples

$\int x^{3} \cos (x) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x^{3}$ and $d v = \cos (x)$ .

$x^{3} \sin (x) - \int \sin (x) (3 x^{2}) d x$

Step 2

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$x^{3} \sin (x) - (3 \int \sin (x) (x^{2}) d x)$

Step 3

Multiply $3$ by $- 1$ .

$x^{3} \sin (x) - 3 \int \sin (x) (x^{2}) d x$

Step 4

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x^{2}$ and $d v = \sin (x)$ .

$x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) - \int - \cos (x) (2 x) d x)$

Step 5

Multiply $2$ by $- 1$ .

$x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) - \int - 2 \cos (x) x d x)$

Step 6

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

$x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) - (- 2 \int \cos (x) x d x))$

Step 7

Multiply $- 2$ by $- 1$ .

$x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) + 2 \int \cos (x) x d x)$

Step 8

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \cos (x)$ .

$x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) + 2 (x \sin (x) - \int \sin (x) d x))$

Step 9

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

$x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) + 2 (x \sin (x) - (- \cos (x) + C)))$

Step 10

Simplify the answer.

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Step 10.1

Rewrite $x^{3} \sin (x) - 3 (x^{2} (- \cos (x)) + 2 (x \sin (x) - (- \cos (x) + C)))$ as $x^{3} \sin (x) - 3 (- x^{2} \cos (x) + 2 x \sin (x) + 2 \cos (x)) + C$ .

$x^{3} \sin (x) - 3 (- x^{2} \cos (x) + 2 x \sin (x) + 2 \cos (x)) + C$

Step 10.2

Rewrite $x^{3} \sin (x) - 3 (- x^{2} \cos (x) + 2 x \sin (x) + 2 \cos (x)) + C$ as $x^{3} \sin (x) + 3 (x^{2} \cos (x)) - 6 (x \sin (x)) - 6 \cos (x) + C$ .

$x^{3} \sin (x) + 3 (x^{2} \cos (x)) - 6 (x \sin (x)) - 6 \cos (x) + C$

Step 10.3

Remove unnecessary parentheses.

$x^{3} \sin (x) + 3 x^{2} \cos (x) - 6 x \sin (x) - 6 \cos (x) + C$