What is the integral of (cos(x))/(sin^2(x)-2sin(x)-8) ?

The integral of (cos(x))/(sin^2(x)-2sin(x)-8) is -1/6 ln|sin(x)+2|+1/6 ln|sin(x)-4|+C

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integral of (cos(x))/(sin^2(x)-2sin(x)-8)

\int \frac{cos ( x )}{sin ^{2} ( x ) - 2 sin ( x ) - 8} d x

- \frac{1}{6} ln ∣ sin (x) + 2 ∣ + \frac{1}{6} ln ∣ sin (x) - 4 ∣ + C

Solution steps

\int \frac{cos ( x )}{sin ^{2} ( x ) - 2 sin ( x ) - 8} d x

Factor

\frac{cos ( x )}{sin ^{2} ( x ) - 2 sin ( x ) - 8}

= \int \frac{cos ( x )}{( sin ( x ) + 2 ) ( sin ( x ) - 4 )} d x

Apply u-substitution

: \int \frac{1}{( u + 2 ) ( u - 4 )} d u

= \int \frac{1}{( u + 2 ) ( u - 4 )} d u

Take the partial fraction of

\frac{1}{( u + 2 ) ( u - 4 )} : - \frac{1}{6 ( u + 2 )} + \frac{1}{6 ( u - 4 )}

= \int - \frac{1}{6 ( u + 2 )} + \frac{1}{6 ( u - 4 )} 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}{6 ( u + 2 )} d u + \int \frac{1}{6 ( u - 4 )} d u

\int \frac{1}{6 ( u + 2 )} d u = \frac{1}{6} ln ∣ u + 2 ∣

\int \frac{1}{6 ( u - 4 )} d u = \frac{1}{6} ln ∣ u - 4 ∣

= - \frac{1}{6} ln ∣ u + 2 ∣ + \frac{1}{6} ln ∣ u - 4 ∣

Substitute back

u = sin (x)

= - \frac{1}{6} ln ∣ sin (x) + 2 ∣ + \frac{1}{6} ln ∣ sin (x) - 4 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ sin (x) + 2 ∣ + \frac{1}{6} ln ∣ sin (x) - 4 ∣ + C

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What is the integral of (cos(x))/(sin^2(x)-2sin(x)-8) ?
The integral of (cos(x))/(sin^2(x)-2sin(x)-8) is -1/6 ln|sin(x)+2|+1/6 ln|sin(x)-4|+C