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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Apply the Power Rule

= 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

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