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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \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

- 4 t^{2} y^{^{''}} + 5 t y^{^{'}} - 2 y = 0

\int \frac{sec ^{2} ( \frac{1}{x ^{9}} )}{x ^{10}} d x

\int \frac{1}{x \cdot ln ( x ^{6} )} d x

x^{^{''}} + 16 x = 0

\int_{1}^{2} \frac{3 x - 1}{3 x} d x

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