Integrate 3 x 2 − 3 2 x 2 − 3 sin ⁡ ( 3 x ) + 1 2 with respect to x.

Question

Integrate   3      x          2            ﻿    −      3          2              x                  2                      −  3  sin  ⁡  (  3  x  )  +      1    2   with respect to x.

Ari Jacobs · Accepted Answer

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

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

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

Simplify the expression.

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

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

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

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

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

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

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

Combine $\sin (u)$ and $\frac{1}{3}$ .

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

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

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

Simplify.

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

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

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

Apply the constant rule.

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

Simplify.

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

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

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

Integrate 3 x 2 </mrow>  </mrow> − 3 <mr

Answered question

Answer & Explanation

New Questions in Integral Calculus