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

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

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

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

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

Solution steps

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

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{y}{2} ) + 1})

Simplify

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

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

Add a constant to the solution

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

y = x^{3}, x = y^{2} - 2

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 5)

\int \frac{7}{( x - 1 ) x ^{2} - 2 x - 48} d x

f (x) = \frac{1}{x ^{4}}

\int arccot (12 y) d y

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