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

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

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

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

sin (x) + x + C

Solution steps

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

Rewrite using trig identities

= \int cos (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 cos (x) d x + \int 1 d x

\int cos (x) d x = sin (x)

\int 1 d x = x

= sin (x) + x

Add a constant to the solution

= sin (x) + x + C

\frac{d}{d x} (arcsin (9 x^{5}))

\frac{8 e ^{x}}{9 e ^{x} + 5}

\int e^{x} sin (yz) d x

\int \frac{x - 1}{x ^{2} - 2 x + 4} d x

\frac{d y}{d x} = \frac{9 x}{10 y}

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