What is the integral of (cos(x))/(2sin^2(x)+3sin(x)) ?

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

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integral of (cos(x))/(2sin^2(x)+3sin(x))

\int \frac{cos ( x )}{2 sin ^{2} ( x ) + 3 sin ( x )} d x

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

Solution steps

\int \frac{cos ( x )}{2 sin ^{2} ( x ) + 3 sin ( x )} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 3 u} d u

= \int \frac{1}{2 u ^{2} + 3 u} d u

Take the partial fraction of

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

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

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

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

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

Substitute back

u = sin (x)

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

Add a constant to the solution

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

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