Question

Evaluate the integral of cos(x) / 3-sin(x)

I'm stuck...

Answer 1

i'll type my working out

Answer 2

did you try the u substitution as
\(\large u= 3-\sin x\)

Answer 3

yep

Answer 4

du =... ?

Answer 5

du/d = -cos(x)

Answer 6

dx = du / -cos(x)

Answer 7

so,
-du = cos x dx
right ?

Answer 8

don't you solve for dx though?

Answer 9

you can...not necessary...

Answer 10

u = 3-sin(x)
\[\frac{ du }{ dx }=-\cos(x)\]
\[dx = \frac{ du }{ -\cos(x) }\]

Answer 11

well that's the way amistre showed me and also I copied some notes in a lecture today which follows that way as well

Answer 12

yes, do that then, and tell me what yur new integral is

Answer 13

\[\int\limits_{}^{}-\frac{ 1 }{ u }du\]

Answer 14

correct,
know the standard formula for that ?

Answer 15

that's after i substitute the u and cancel out the cos(x)

Answer 16

nope, I'm stuck at that part :(

Answer 17

wait is it -log(u) + C?

Answer 18

\(\large \int \dfrac{f'(x)}{f(x)} dx= \log f(x)+c \\ \large \int \dfrac{1}{x} dx= \log x+c\)

yes!

Answer 19

so the final answer should be -log(3-sin(x))+C i believe?

Answer 20

correct! :)

Answer 21

k thanks!

Answer 22

welcome ^_^