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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

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

Substitute back

u = cos (x)

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

Simplify

= \frac{1}{3} ln ∣ cos (x) ∣ - \frac{1}{3} ln ∣ cos (x) - 3 ∣

Add a constant to the solution

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

f (x) = x^{2} - x + 4, at x = 4

\int (8 t^{4} + 5 t^{3} - 6 t^{2} + 7 t - 9) d t

\frac{( s + 3 )}{( s ^{2} + s + 3 )}

\frac{d}{d x} (- e^{- x} sin (x) + e^{- x} cos (x))

\frac{\partial}{\partial x} (y z^{2} cos (x))

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