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

The integral of (1-sin(x))/(1+sin(x)) is 2(-2/(tan(x/2)+1)-arctan(tan(x/2)))+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

= 2 (- \frac{2}{u + 1} - arctan (u))

Substitute back

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

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

Add a constant to the solution

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

\frac{d r}{d θ} + r tan (θ) = 5 tan (θ)

y^{^{''''}} + y^{^{''}} - 6 y = 0

f (x) = sin (2 x)

\int t^{3} sin (n t) d t

\int - \frac{2}{x ^{2} - 1} d x

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