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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Complete the square

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

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\frac{( 3 y ^{2} - t ^{2} )}{( y ^{5} )} \cdot y^{^{'}} + \frac{t}{( 2 y ^{4} )} = 0, y (1) = 3

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