Question

Integration by parts

Answer 1

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

Answer 2

you dont need IBP

Answer 3

what?

Answer 4

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

Answer 5

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

Answer 6

I don't understand the basic formula

Answer 7

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

Answer 8

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 \(-10\)

Answer 9

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