What is the integral of (sin(x))/((1+sin(x))^2) ?

The integral of (sin(x))/((1+sin(x))^2) is 4(-1/(2(tan(x/2)+1)^2)+1/(3(tan(x/2)+1)^3))+C

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integral of (sin(x))/((1+sin(x))^2)

\int \frac{sin ( x )}{( 1 + sin ( x ) ) ^{2}} d x

4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{3 ( tan ( \frac{x}{2} ) + 1 ) ^{3}}) + C

Solution steps

\int \frac{sin ( x )}{( 1 + sin ( x ) ) ^{2}} d x

Apply u-substitution

: \int \frac{4 u}{( 1 + u ^{2} + 2 u ) ^{2}} d u

= \int \frac{4 u}{( 1 + u ^{2} + 2 u ) ^{2}} d u

Take the constant out:

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

= 4 \cdot \int \frac{u}{( 1 + u ^{2} + 2 u ) ^{2}} d u

Take the partial fraction of

\frac{u}{( 1 + u ^{2} + 2 u ) ^{2}} : \frac{1}{( u + 1 ) ^{3}} - \frac{1}{( u + 1 ) ^{4}}

= 4 \cdot \int \frac{1}{( u + 1 ) ^{3}} - \frac{1}{( u + 1 ) ^{4}} 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

= 4 (\int \frac{1}{( u + 1 ) ^{3}} d u - \int \frac{1}{( u + 1 ) ^{4}} d u)

\int \frac{1}{( u + 1 ) ^{3}} d u = - \frac{1}{2 ( u + 1 ) ^{2}}

\int \frac{1}{( u + 1 ) ^{4}} d u = - \frac{1}{3 ( u + 1 ) ^{3}}

= 4 (- \frac{1}{2 ( u + 1 ) ^{2}} - (- \frac{1}{3 ( u + 1 ) ^{3}}))

Substitute back

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

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} - (- \frac{1}{3 ( tan ( \frac{x}{2} ) + 1 ) ^{3}}))

Simplify

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{3 ( tan ( \frac{x}{2} ) + 1 ) ^{3}})

Add a constant to the solution

= 4 (- \frac{1}{2 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{3 ( tan ( \frac{x}{2} ) + 1 ) ^{3}}) + C

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What is the integral of (sin(x))/((1+sin(x))^2) ?
The integral of (sin(x))/((1+sin(x))^2) is 4(-1/(2(tan(x/2)+1)^2)+1/(3(tan(x/2)+1)^3))+C