What is the integral of 1/((1-sin(x))^2) ?

The integral of 1/((1-sin(x))^2) is 2(-1/(tan(x/2)-1)-1/((tan(x/2)-1)^2)-2/(3(tan(x/2)-1)^3))+C

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integral of 1/((1-sin(x))^2)

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

2 (- \frac{1}{tan ( \frac{x}{2} ) - 1} - \frac{1}{( tan ( \frac{x}{2} ) - 1 ) ^{2}} - \frac{2}{3 ( tan ( \frac{x}{2} ) - 1 ) ^{3}}) + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{u ^{2} + 1}{( 1 + u ^{2} - 2 u ) ^{2}} : \frac{1}{( u - 1 ) ^{2}} + \frac{2}{( u - 1 ) ^{3}} + \frac{2}{( u - 1 ) ^{4}}

= 2 \cdot \int \frac{1}{( u - 1 ) ^{2}} + \frac{2}{( u - 1 ) ^{3}} + \frac{2}{( u - 1 ) ^{4}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{( u - 1 ) ^{2}} d u + \int \frac{2}{( u - 1 ) ^{3}} d u + \int \frac{2}{( u - 1 ) ^{4}} d u)

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

\int \frac{2}{( u - 1 ) ^{3}} d u = - \frac{1}{( u - 1 ) ^{2}}

\int \frac{2}{( u - 1 ) ^{4}} d u = - \frac{2}{3 ( u - 1 ) ^{3}}

= 2 (- \frac{1}{u - 1} - \frac{1}{( u - 1 ) ^{2}} - \frac{2}{3 ( u - 1 ) ^{3}})

Substitute back

u = tan (\frac{x}{2})

= 2 (- \frac{1}{tan ( \frac{x}{2} ) - 1} - \frac{1}{( tan ( \frac{x}{2} ) - 1 ) ^{2}} - \frac{2}{3 ( tan ( \frac{x}{2} ) - 1 ) ^{3}})

Add a constant to the solution

= 2 (- \frac{1}{tan ( \frac{x}{2} ) - 1} - \frac{1}{( tan ( \frac{x}{2} ) - 1 ) ^{2}} - \frac{2}{3 ( tan ( \frac{x}{2} ) - 1 ) ^{3}}) + C

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What is the integral of 1/((1-sin(x))^2) ?
The integral of 1/((1-sin(x))^2) is 2(-1/(tan(x/2)-1)-1/((tan(x/2)-1)^2)-2/(3(tan(x/2)-1)^3))+C