Mathematics OpenStudy (anonymous):

Evaluate the indefinite integral cos(x)/7sin(x)+35 OpenStudy (amistre64):

$\frac{cos(x)}{7sin(x)}+35$ this? OpenStudy (amistre64):

or is the 35 tucked under as well? OpenStudy (anonymous):

the 35 is tucked under as well OpenStudy (amistre64):

$\frac{1}{7}\frac{cos}{sin+5}$is the same then correct? OpenStudy (anonymous):

yes myininaya (myininaya):

let u=denominator du= 7cos(x) dx OpenStudy (amistre64):

notice that the top is the derivative of the bottom; so we get a 1/x type setup OpenStudy (amistre64):

$\frac{1}{7}\int\frac{Dx(sin+5)}{sin+5}=?$ OpenStudy (anonymous):

ok so i will get $1/7(\sin(x))+5$? OpenStudy (amistre64):

recall that:$Dx(ln(x))=\frac{Dx}{x}$ OpenStudy (amistre64):

or some such notation :) OpenStudy (anonymous):

so 1/7(ln(x))+5 I'm still a little lost OpenStudy (amistre64): OpenStudy (amistre64):

let me try this way OpenStudy (amistre64):

$\frac{d}{dx}({ln(x)})=\frac{dx}{x}\text{; where dx is the derivative of "x"}$ $\frac{d}{dx}(sin(x)+5)=cos(x)$ $\int \frac{cos(x)}{sin(x)+5}\text{ is similar to form as }\int \frac{dx}{x}$ $\int \frac{dx}{x+5}=ln|x+5|+C$ $\int \frac{cos(x)}{sin(x)+5}=ln|sin(x)+5|+C$ OpenStudy (amistre64):

but lets not forget the 1/7 we took out OpenStudy (amistre64):

$\frac{1}{7}ln|sin(x)+5|\ +C$ OpenStudy (anonymous):

ok i got it. Thank you so much OpenStudy (amistre64):

youre welcome, just gotta recall those derivative rules :)

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