What is the integral of 1/((2+sin(x))cos(x)) ?

The integral of 1/((2+sin(x))cos(x)) is 1/2 ln|sin(x)+1|-1/6 ln|sin(x)-1|-1/3 ln|sin(x)+2|+C

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integral of 1/((2+sin(x))cos(x))

\int \frac{1}{( 2 + sin ( x ) ) cos ( x )} d x

\frac{1}{2} ln ∣ sin (x) + 1 ∣ - \frac{1}{6} ln ∣ sin (x) - 1 ∣ - \frac{1}{3} ln ∣ sin (x) + 2 ∣ + C

Solution steps

\int \frac{1}{( 2 + sin ( x ) ) cos ( x )} d x

Apply u-substitution

: \int \frac{1}{( 1 - u ^{2} ) ( 2 + u )} d u

= \int \frac{1}{( 1 - u ^{2} ) ( 2 + u )} d u

Take the partial fraction of

\frac{1}{( 1 - u ^{2} ) ( 2 + u )} : \frac{1}{2 ( u + 1 )} - \frac{1}{6 ( u - 1 )} - \frac{1}{3 ( u + 2 )}

= \int \frac{1}{2 ( u + 1 )} - \frac{1}{6 ( u - 1 )} - \frac{1}{3 ( u + 2 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 ( u + 1 )} d u - \int \frac{1}{6 ( u - 1 )} d u - \int \frac{1}{3 ( u + 2 )} d u

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

\int \frac{1}{6 ( u - 1 )} d u = \frac{1}{6} ln ∣ u - 1 ∣

\int \frac{1}{3 ( u + 2 )} d u = \frac{1}{3} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{6} ln ∣ u - 1 ∣ - \frac{1}{3} ln ∣ u + 2 ∣

Substitute back

u = sin (x)

= \frac{1}{2} ln ∣ sin (x) + 1 ∣ - \frac{1}{6} ln ∣ sin (x) - 1 ∣ - \frac{1}{3} ln ∣ sin (x) + 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ sin (x) + 1 ∣ - \frac{1}{6} ln ∣ sin (x) - 1 ∣ - \frac{1}{3} ln ∣ sin (x) + 2 ∣ + C

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What is the integral of 1/((2+sin(x))cos(x)) ?
The integral of 1/((2+sin(x))cos(x)) is 1/2 ln|sin(x)+1|-1/6 ln|sin(x)-1|-1/3 ln|sin(x)+2|+C