What is the integral of 1/(csc^4(y/2)) ?

The integral of 1/(csc^4(y/2)) is 2(-1/4 sin^3(y/2)cos(y/2)+3/16 (y-sin(y)))+C

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integral of 1/(csc^4(y/2))

\int \frac{1}{csc ^{4} ( \frac{y}{2} )} d y

2 (- \frac{1}{4} sin^{3} (\frac{y}{2}) cos (\frac{y}{2}) + \frac{3}{16} (y - sin (y))) + C

Solution steps

\int \frac{1}{csc ^{4} ( \frac{y}{2} )} d y

Use the basic trigonometric identity:

\frac{1}{csc ( x )} = sin (x)

= \int sin^{4} (\frac{y}{2}) d y

Apply u-substitution

= \int sin^{4} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int sin^{4} (u) d u

Apply Integral Reduction

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

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

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

Substitute back

u = \frac{y}{2}

= 2 (- \frac{cos ( \frac{y}{2} ) sin ^{3} ( \frac{y}{2} )}{4} + \frac{3}{4} \cdot \frac{1}{2} (\frac{y}{2} - \frac{1}{2} sin (2 \cdot \frac{y}{2})))

Simplify

2 (- \frac{cos ( \frac{y}{2} ) sin ^{3} ( \frac{y}{2} )}{4} + \frac{3}{4} \cdot \frac{1}{2} (\frac{y}{2} - \frac{1}{2} sin (2 \cdot \frac{y}{2}))) : 2 (- \frac{1}{4} sin^{3} (\frac{y}{2}) cos (\frac{y}{2}) + \frac{3}{16} (y - sin (y)))

= 2 (- \frac{1}{4} sin^{3} (\frac{y}{2}) cos (\frac{y}{2}) + \frac{3}{16} (y - sin (y)))

Add a constant to the solution

= 2 (- \frac{1}{4} sin^{3} (\frac{y}{2}) cos (\frac{y}{2}) + \frac{3}{16} (y - sin (y))) + C

What is the integral of 1/(csc^4(y/2)) ?
The integral of 1/(csc^4(y/2)) is 2(-1/4 sin^3(y/2)cos(y/2)+3/16 (y-sin(y)))+C