What is the integral of 2/(-sin(t)+1) ?

The integral of 2/(-sin(t)+1) is -4/(tan(t/2)-1)+C

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integral of 2/(-sin(t)+1)

\int \frac{2}{- sin ( t ) + 1} d t

- \frac{4}{tan ( \frac{t}{2} ) - 1} + C

Solution steps

\int \frac{2}{- sin ( t ) + 1} d t

Take the constant out:

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

= 2 \cdot \int \frac{1}{- sin ( t ) + 1} d t

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

= 2 \cdot 2 (- \frac{1}{tan ( \frac{t}{2} ) - 1})

Simplify

2 \cdot 2 (- \frac{1}{tan ( \frac{t}{2} ) - 1}) : - \frac{4}{tan ( \frac{t}{2} ) - 1}

= - \frac{4}{tan ( \frac{t}{2} ) - 1}

Add a constant to the solution

= - \frac{4}{tan ( \frac{t}{2} ) - 1} + C

\frac{d}{d x} (\frac{1}{lo g _{e} ( x )})

\frac{d y}{d x} + 3 y = e^{(- 5 x)}

\int \frac{3 x ^{3}}{x ^{3} - 4 x} d x

(6 n - 1) (- 2 x)^{n}

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

What is the integral of 2/(-sin(t)+1) ?
The integral of 2/(-sin(t)+1) is -4/(tan(t/2)-1)+C