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

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

or is the 35 tucked under as well?

the 35 is tucked under as well

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

yes

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

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

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

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

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

or some such notation :)

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

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

let me try this way

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

but lets not forget the 1/7 we took out

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

ok i got it. Thank you so much

youre welcome, just gotta recall those derivative rules :)

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