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

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

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

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

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

Solution steps

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

\frac{1 + sin ( x )}{1 - sin ( x )} = \frac{2}{1 - sin ( x )} - 1

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

Apply the Sum Rule:

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

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

\int \frac{2}{1 - sin ( x )} d x = - \frac{4}{tan ( \frac{x}{2} ) - 1}

\int 1 d x = x

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

Add a constant to the solution

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

\int - \frac{1}{2 t} d t

\int (- 2)^{x} d x

\int (1 - x)^{10} d x

\int \frac{x + 2}{( x - 4 ) ^{2}} d x

\int \frac{2}{x} + 1 d x

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