Question

take the anti-derivative of

sin(1/7x) dx

Answer 1

i know that we have to use the chain rule

Answer 2

\[\int\limits \sin(\frac{ 1 }{ 7 }x)dx\]

Answer 3

backwards

Answer 4

since it sin it would have to go to -cos

Answer 5

you want to say \(-\cos(\frac{x}{7})\) but that is wrong because you would get
\[\frac{1}{7}\sin(\frac{x}{7})\] by the chain rule
no worries, multiply by 7 to fix is

Answer 6

that what i dont understant. where is the extra 7 coming from

Answer 7

let u = 1/7 x
du = 1/7 dx so dx = 7 du

\[\int\limits \sin(u) *7 *du\]

integrate, then sub back in what x is

Answer 8

this is substitution right, i didnt think you could use it for this one

Answer 9

yeah, u - sub, so you get

\[7*\int\limits \sin(u) du = 7*-\cos(u) \]

u = 1/7 x

Answer 10

-7 cos(1/7 x)

Answer 11

then you pulled the 7 out because it was a constant right

Answer 12

right, you let u = 1/7x and the derivative is du = 1/7 dx

solve that for dx, dx = 7 du

pull the constant outside the integral

Answer 13

the extra 7 from the chain rule
if you take the derivative you have to divide by 7, so when you take the anti derivative you have to multiply by 7

Answer 14

-7cos(x/7) + c

Answer 15

okay why did i leave the du alone in the equation for example

7/ sin(u) du

Answer 16

wouldnt i have to take it back to dx or no

Answer 17

you make a substitution, so your integral is in U-space , you complete the integral in U-space, then substitute back in at the end

Answer 18

you defined u = 1/7 x

Answer 19

Okay i get that, thank you so much

Answer 20

welcome