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

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

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

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

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

Solution steps

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

Expand

\frac{1 - cos ( x )}{csc ^{2} ( x )} : \frac{1}{csc ^{2} ( x )} - \frac{cos ( x )}{csc ^{2} ( x )}

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

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

\int \frac{cos ( x )}{csc ^{2} ( x )} d x = \frac{sin ^{3} ( x )}{3}

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

Add a constant to the solution

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

\frac{d}{d x} ((8 x + 3)^{2})

\frac{d}{d x} ((3 x^{4} + 1)^{3})

\int sec^{2} (x) tan (x) d x

\frac{d}{d x} (\frac{C ^{2} - x ^{2}}{2 x})

y^{^{''}} = x, y (- 1) = 0, y (1) = 0

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