What is the integral of (sin(7x))/(3cos(7x)+13) ?

The integral of (sin(7x))/(3cos(7x)+13) is -1/21 ln|3cos(7x)+13|+C

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integral of (sin(7x))/(3cos(7x)+13)

\int \frac{sin ( 7 x )}{3 cos ( 7 x ) + 13} d x

- \frac{1}{21} ln ∣ 3 cos (7 x) + 13 ∣ + C

Solution steps

\int \frac{sin ( 7 x )}{3 cos ( 7 x ) + 13} d x

Apply u-substitution

= \int - \frac{1}{21 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{21} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{21} ln ∣ u ∣

Substitute back

u = 3 cos (7 x) + 13

= - \frac{1}{21} ln ∣ 3 cos (7 x) + 13 ∣

Add a constant to the solution

= - \frac{1}{21} ln ∣ 3 cos (7 x) + 13 ∣ + C

What is the integral of (sin(7x))/(3cos(7x)+13) ?
The integral of (sin(7x))/(3cos(7x)+13) is -1/21 ln|3cos(7x)+13|+C