To solve the integral [tex]\(\int -14 x \sin(7x^2 - 8) \, dx\)[/tex] using the substitution [tex]\(u = 7x^2 - 8\)[/tex], follow these steps:
1. Substitute and find the differential in terms of [tex]\(dx\)[/tex]:
Let [tex]\(u = 7x^2 - 8\)[/tex].
Then, differentiate [tex]\(u\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{du}{dx} = 14x
\][/tex]
Hence,
[tex]\[
du = 14x \, dx
\][/tex]
Solving for [tex]\(dx\)[/tex], we get:
[tex]\[
dx = \frac{du}{14x}
\][/tex]
2. Rewrite the integral in terms of [tex]\(u\)[/tex]:
Substitute [tex]\(u = 7x^2 - 8\)[/tex] and [tex]\(dx = \frac{du}{14x}\)[/tex] into the integral:
[tex]\[
\int -14x \sin(7x^2 - 8) \, dx = \int -14x \sin(u) \cdot \frac{du}{14x}
\][/tex]
Notice that the [tex]\(14x\)[/tex] terms cancel out:
[tex]\[
\int -14x \sin(u) \cdot \frac{du}{14x} = \int -\sin(u) \, du
\][/tex]
3. Integrate with respect to [tex]\(u\)[/tex]:
The integral [tex]\(\int -\sin(u) \, du\)[/tex] is a standard integral in calculus:
[tex]\[
\int -\sin(u) \, du = \cos(u) + C
\][/tex]
4. Substitute back in terms of [tex]\(x\)[/tex]:
Recall the substitution [tex]\(u = 7x^2 - 8\)[/tex]:
[tex]\[
\cos(u) + C = \cos(7x^2 - 8) + C
\][/tex]
Thus, the indefinite integral is:
[tex]\[
\int -14x \sin(7x^2 - 8) \, dx = \cos(7x^2 - 8) + C
\][/tex]
Check by Differentiation:
To verify our answer, differentiate [tex]\(\cos(7x^2 - 8) + C\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d}{dx} \left(\cos(7x^2 - 8) + C\right)
\][/tex]
Using the chain rule:
[tex]\[
\frac{d}{dx} \cos(7x^2 - 8) = -\sin(7x^2 - 8) \cdot \frac{d}{dx} (7x^2 - 8)
\][/tex]
Since [tex]\(\frac{d}{dx} (7x^2 - 8) = 14x\)[/tex], we have:
[tex]\[
-\sin(7x^2 - 8) \cdot 14x = -14x \sin(7x^2 - 8)
\][/tex]
This matches the original integrand. Therefore, our integration was done correctly.
The answer is:
[tex]\[
\int -14x \sin(7x^2 - 8) \, dx = \cos(7x^2 - 8) + C
\][/tex]
