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

Question

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

Ayaan Gonzalez · Accepted Answer

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

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

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

Simplify the expression.

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

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

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

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

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

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

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

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

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

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

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

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

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

Apply the constant rule.

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

Simplify.

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

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

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

Reorder terms.

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

Integrate − 3 x 2 </mrow> − 2 x