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

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

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integral of sqrt((1+cos(x))/(1-cos(x)))

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 2 (ln ∣ u ∣ - \frac{1}{2} ln u^{2} + 1)

Substitute back

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

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

Simplify

2 (ln tan (\frac{x}{2}) - \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1) : 2 ln tan (\frac{x}{2}) - ln (sec^{2} (\frac{x}{2}))

= 2 ln tan (\frac{x}{2}) - ln (sec^{2} (\frac{x}{2}))

Add a constant to the solution

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

\frac{d}{d x} (arcsin (5 x))

\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}} + 9

x \to - 2 + lim (\frac{1 - x}{x ^{3} + 8})

x \to 1 - lim (\frac{2}{x - 1})

\frac{\partial}{\partial x} ((3 x^{2} + y) e^{x y^{2}})

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