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

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

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integral of 1/(sin(u)+1)

\int \frac{1}{sin ( u ) + 1} d u

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

Solution steps

\int \frac{1}{sin ( u ) + 1} d u

Apply u-substitution

: \int \frac{2}{2 v + 1 + v ^{2}} d v

= \int \frac{2}{2 v + 1 + v ^{2}} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{2 v + 1 + v ^{2}} d v

Complete the square

2 v + 1 + v^{2} : (v + 1)^{2}

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

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

= 2 \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= 2 (- \frac{1}{w})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = \frac{x ^{2} y ^{4} + 3 x ^{2}}{y ^{3}}

n = 1 \sum \infty \frac{1}{n} (\frac{3}{2})^{n}

x \to 4 lim (\frac{7 x - 1}{x - 4})

\frac{d}{d x} (3 e^{4 x^{2}})

\frac{s}{( s ^{2} + 9 ) \cdot ( s ^{2} + 16 )}

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