Integration by parts

Question

anonymous · Answer

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

mimi_x3 · Answer

you  dont need IBP

anonymous · Answer

what?

mimi_x3 · Answer

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

raden · Answer

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

anonymous · Answer

I don't understand the basic formula

anonymous · Answer

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

anonymous · Answer

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 $-10$

anonymous · Answer

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