## Brooke_army Group Title Integration by parts one year ago one year ago

1. Brooke_army Group Title

$\int\limits_{}^{} \cos(-10x)dx$ u=-10x

2. Mimi_x3 Group Title

you dont need IBP

3. Brooke_army Group Title

what?

4. Mimi_x3 Group Title

$\int\limits cosaxdx =\frac{1}{a} sinax+c$

just use the basic formula : int (cosAx) dx = 1/A * sinAx + c

6. Brooke_army Group Title

I don't understand the basic formula

7. satellite73 Group Title

@Brooke_army "parts" is something else that you may not have seen yet. this type of integration is usually called "u - substitution"

8. satellite73 Group Title

if you didn't have the $$-10x$$ you would have $\int cos(x)dx=\sin(x)$ because $\frac{d}{dx}[\sin(x)]=\cos(x)$ but $\frac{d}{dx}[\sin(-10x)]=-10\cos(-10x)$ by the chain rule therefore, to get what you want, you have to divide by $$-10$$

9. satellite73 Group Title

so if you take the derivative of $$-\frac{1}{10}\sin(-10x)$$ you get exactly what you want, namely $$\cos(-10x)$$