What is the integral of (cos(x))/(9-sin(x)) ?

The integral of (cos(x))/(9-sin(x)) is -ln|9-sin(x)|+C

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integral of (cos(x))/(9-sin(x))

\int \frac{cos ( x )}{9 - sin ( x )} d x

- ln ∣ 9 - sin (x) ∣ + C

Solution steps

\int \frac{cos ( x )}{9 - sin ( x )} d x

Apply u-substitution

: \int - \frac{1}{u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 9 - sin (x)

= - ln ∣ 9 - sin (x) ∣

Add a constant to the solution

= - ln ∣ 9 - sin (x) ∣ + C

\frac{d}{d x} (3 x (10 x^{3} - \frac{6}{x + 10}))

\frac{d}{d x} (- 5 cos (5 x))

x \to - \infty lim (5^{x} + 2)

\int_{0}^{\frac{1}{2}} 3 arccos (x) d x

\frac{\partial}{\partial y} (x \cdot ln (y - z))

What is the integral of (cos(x))/(9-sin(x)) ?
The integral of (cos(x))/(9-sin(x)) is -ln|9-sin(x)|+C