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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{- 2 y ^{2}}{x ^{2}}

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 7)

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

\frac{d}{d x} (e^{x y z})

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