Question

Evaluate the indefinite integral cos(x)/7sin(x)+35

Answer 1

\[\frac{cos(x)}{7sin(x)}+35\] this?

Answer 2

or is the 35 tucked under as well?

Answer 3

the 35 is tucked under as well

Answer 4

\[\frac{1}{7}\frac{cos}{sin+5}\]is the same then correct?

Answer 5

yes

Answer 6

let u=denominator
du= 7cos(x) dx

Answer 7

notice that the top is the derivative of the bottom; so we get a 1/x type setup

Answer 8

\[\frac{1}{7}\int\frac{Dx(sin+5)}{sin+5}=?\]

Answer 9

ok so i will get \[1/7(\sin(x))+5\]?

Answer 10

recall that:\[Dx(ln(x))=\frac{Dx}{x}\]

Answer 11

or some such notation :)

Answer 12

so 1/7(ln(x))+5 I'm still a little lost

Answer 13

your getting close, but your form is still a bit off

Answer 14

let me try this way

Answer 15

\[\frac{d}{dx}({ln(x)})=\frac{dx}{x}\text{; where dx is the derivative of "x"}\]
\[\frac{d}{dx}(sin(x)+5)=cos(x)\]
\[\int \frac{cos(x)}{sin(x)+5}\text{ is similar to form as }\int \frac{dx}{x}\]
\[\int \frac{dx}{x+5}=ln|x+5|+C\]
\[\int \frac{cos(x)}{sin(x)+5}=ln|sin(x)+5|+C\]

Answer 16

but lets not forget the 1/7 we took out

Answer 17

\[\frac{1}{7}ln|sin(x)+5|\ +C\]

Answer 18

ok i got it. Thank you so much

Answer 19

youre welcome, just gotta recall those derivative rules :)