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

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

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integral of (cos(x))/(sin(x)(1-sin(x)))

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

- 2 ln tan (\frac{x}{2}) - 1 + ln tan (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{u + 1}{u ( u - 1 )} : \frac{u}{u ( u - 1 )} + \frac{1}{u ( u - 1 )}

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

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

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

= - (ln ∣ u - 1 ∣ - ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = tan (\frac{x}{2})

= - (ln tan (\frac{x}{2}) - 1 - ln tan (\frac{x}{2}) + ln tan (\frac{x}{2}) - 1)

Simplify

= - 2 ln tan (\frac{x}{2}) - 1 + ln tan (\frac{x}{2})

Add a constant to the solution

= - 2 ln tan (\frac{x}{2}) - 1 + ln tan (\frac{x}{2}) + C

\frac{d}{d t} (\frac{v ( t )}{t})

cos^{3} (sin (2 x^{4} - x))

y = x^{2} - 7, y = 9

\int e^{- in wt} d t

\frac{d}{d x} (- 3 \cdot e^{x^{2} + x - 1})

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