Question

How to integrate the following question?

Answer 1

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Answer 2

\[\int\limits6x \sin (x ^{2}-4) dx \] given u= x^2 - 4

Answer 3

take the derivative of that

Answer 4

u, what do you get?

Answer 5

du/dx = 2x

Answer 6

alright so you know you can pull constants out of an integral correct, if you pull 3 out, what do you get?

Answer 7

uh 2xsin)x^2-4) ??

Answer 8

yep so back to du/dx

if you multiply by dx

you get

du=2xdx... which you have so you can replace it with du

\[3\int{}{}sin(u)du\]

Answer 9

now integrate and replace u with your function and don't forget + c =]

Answer 10

Thank you! so I ought to get \[3 \int\limits 2x \sin (x ^{2} - 4) dx\] do I hace to integrate the sin as well? into -cos?

Answer 11

yes the integral of sin(u) = -cos(u) you don't really have to pull out the 3 you can keep it there it's just i'm used to dealing with du rather than dx

if you didn't pull out the three you'd get

u=x^2-4
du/dx=2x

du=2xdx

du/2x=dx

\[\int{}{}\frac{6xsin(x^2-4)}{2x}dx\]

the x's would cancel and 6/2 equal 3 so you'd get

\[\int{}{}3sin(x^2-4)dx\]

Answer 12

however i like u better since most books give integration rules in u substitutions

Answer 13

integrating you should get

-3cos(u)+c

then substitute your u back into it to get

-3cos(x^2-4)+c

Answer 14

yes, even my teacher always give it in u substitutions. Alright I'll try to work it out ! Thank you so much for your help! I suck in integration :P

Answer 15

the more you see of it the better you get at it.. You do a loooott of integration after calc 1