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

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

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

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

- ln ∣ cos (x) ∣ + ln ∣ cos (x) - 1 ∣ + C

Solution steps

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

Factor

cos (x) - cos^{2} (x) : - (- cos (x) + cos^{2} (x))

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{- u + u ^{2}} : - \frac{1}{u} + \frac{1}{u - 1}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - (- (- ln ∣ u ∣ + ln ∣ u - 1 ∣))

Substitute back

u = cos (x)

= - (- (- ln ∣ cos (x) ∣ + ln ∣ cos (x) - 1 ∣))

Simplify

= - ln ∣ cos (x) ∣ + ln ∣ cos (x) - 1 ∣

Add a constant to the solution

= - ln ∣ cos (x) ∣ + ln ∣ cos (x) - 1 ∣ + C

\frac{d y}{d x} = \frac{( 1 - y )}{1 - x}

\int 4 cos (2 θ - 5) d θ

\int x + ln (x) d x

x \to 0 + lim (- x + 1)

y^{^{'}} = \frac{y}{x} + (\frac{y}{x})^{2}

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