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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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