Integrate − 3 x 3 2 + x 2 2 − 2 x + 3 sin ⁡ ( 3 x ) with respect to x.

Question

Integrate   −            3              x                  3                      2        ﻿    +            x              2              2    −  2  x  +  3  sin  ⁡  (  3  x  ) with respect to x.

necrologo9yh43 · Accepted Answer

Split the single integral into multiple integrals.

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

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

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

Since $\frac{3}{2}$ is constant with respect to $x$ , move $\frac{3}{2}$ out of the integral.

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

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \int \frac{x^{2}}{2} d x + \int - 2 x d x + \int 3 \sin (3 x) d x$

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

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} \int x^{2} d x + \int - 2 x d x + \int 3 \sin (3 x) d x$

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) + \int - 2 x d x + \int 3 \sin (3 x) d x$

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

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 \int x d x + \int 3 \sin (3 x) d x$

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{1}{2} x^{2} + C) + \int 3 \sin (3 x) d x$

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

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{1}{2} x^{2} + C) + 3 \int \sin (3 x) d x$

Let $u = 3 x$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{1}{2} x^{2} + C) + 3 \int \sin (u) \frac{1}{3} d u$

Simplify.

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{x^{2}}{2} + C) + 3 \int \frac{\sin (u)}{3} d u$

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{x^{2}}{2} + C) + 3 (\frac{1}{3} \int \sin (u) d u)$

Simplify.

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{x^{2}}{2} + C) + \int \sin (u) d u$

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

$- \frac{3}{2} (\frac{1}{4} x^{4} + C) + \frac{1}{2} (\frac{1}{3} x^{3} + C) - 2 (\frac{x^{2}}{2} + C) - \cos (u) + C$

Simplify.

$- \frac{3 x^{4}}{8} + \frac{x^{3}}{6} - x^{2} - \cos (u) + C$

Replace all occurrences of $u$ with $3 x$ .

$- \frac{3 x^{4}}{8} + \frac{x^{3}}{6} - x^{2} - \cos (3 x) + C$

Reorder terms.

$- \frac{3}{8} x^{4} + \frac{1}{6} x^{3} - x^{2} - \cos (3 x) + C$

Integrate − <mrow> 3 x 3 </mrow>

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