Integrate $\int (6-5x^2)(-10x)dx$ using u substitution

Question

Section 5.2

Integrate $\int (6-5x^2)(-10x)dx$ using u substitution

Let $u=6-5x^2$. Then we have:

$\frac{d}{dx}u=\frac{d}{dx}(6-5x^2)$

$\rightarrow \frac{du}{dx}=-10x$

Thus if we make the substitutions in the original integral of $\frac{du}{dx}=-10x$ and $u=6-5x^2$ we get:

$\int(6-5x^2)(-10x)dx=\int u \frac{du}{dx}dx=\int udu$

Where the last equality can be seen as the $dx$ in the denominator canceling with the $dx$ in the numerator. Now that we have simplified things with $u$ substitution, lets integrate:

$\int udu = \frac{u^2}{2}+C=\frac{(6-5x^2)^2}{2}+C$

Answer 1

Actually $\frac{du}{dx}=-10x$, but the rest of what you did is right, including the final answer.

Answer 2

If $u = 6-5x^2$, then $du = -2(5x)\,dx = -10 x\,dx$.

Then you get $$\int u\,du$$

Can you take it from here?? You seem to have proceeded just fine. Perhaps you made a typo in your calculation of $\frac d{dx}$?

Indeed, the integral becomes $$\frac{u^2}2 + c = \frac{(6-5x^2)^2}2 + c$$ after back substitution.