Sure! Let's solve the integrals step by step.
### Part a) [tex]\(\int \frac{1}{x^3} \, dx\)[/tex]
Given the integral to solve:
[tex]\[
\int \frac{1}{x^3} \, dx
\][/tex]
First, we can rewrite [tex]\(\frac{1}{x^3}\)[/tex] in terms of exponent notation:
[tex]\[
\frac{1}{x^3} = x^{-3}
\][/tex]
Now, integrate [tex]\(x^{-3}\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\int x^{-3} \, dx
\][/tex]
To integrate [tex]\(x^n\)[/tex], we use the power rule for integration, which states:
[tex]\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for} \; n \neq -1
\][/tex]
In this case, [tex]\(n = -3\)[/tex]:
[tex]\[
\int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C
\][/tex]
[tex]\[
= \frac{x^{-2}}{-2} + C
\][/tex]
[tex]\[
= -\frac{1}{2} x^{-2} + C
\][/tex]
Rewrite [tex]\(x^{-2}\)[/tex] with a positive exponent:
[tex]\[
= -\frac{1}{2} \cdot \frac{1}{x^2} + C
\][/tex]
[tex]\[
= -\frac{1}{2x^2} + C
\][/tex]
Therefore, the result of the integral is:
[tex]\[
\int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} + C
\][/tex]
### Part b) [tex]\(\int 6 \sin 3x \, dx\)[/tex]
Given the integral to solve:
[tex]\[
\int 6 \sin 3x \, dx
\][/tex]
We can factor out the constant [tex]\(6\)[/tex]:
[tex]\[
6 \int \sin 3x \, dx
\][/tex]
Now, we need to integrate [tex]\(\sin 3x\)[/tex]. We use the substitution method. Let [tex]\(u = 3x\)[/tex], hence [tex]\(du = 3 \, dx\)[/tex] or [tex]\(\frac{du}{3} = dx\)[/tex]:
[tex]\[
6 \int \sin u \cdot \frac{du}{3}
\][/tex]
Factor out the [tex]\(\frac{1}{3}\)[/tex]:
[tex]\[
6 \cdot \frac{1}{3} \int \sin u \, du
\][/tex]
[tex]\[
2 \int \sin u \, du
\][/tex]
Now integrate [tex]\(\sin u\)[/tex]:
[tex]\[
\int \sin u \, du = -\cos u + C
\][/tex]
So:
[tex]\[
2 \left(-\cos u + C\right)
\][/tex]
[tex]\[
= -2 \cos u + 2C
\][/tex]
Substitute back [tex]\(u = 3x\)[/tex]:
[tex]\[
= -2 \cos (3x) + 2C
\][/tex]
Since [tex]\(2C\)[/tex] is just an arbitrary constant, we can denote it simply by [tex]\(C\)[/tex]:
[tex]\[
= -2 \cos (3x) + C
\][/tex]
Therefore, the result of the integral is:
[tex]\[
\int 6 \sin 3x \, dx = -2 \cos (3x) + C
\][/tex]
### Summary of Results
a) [tex]\(\int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} + C\)[/tex]
b) [tex]\(\int 6 \sin 3x \, dx = -2 \cos (3x) + C\)[/tex]
