What is the integral of cot^3(x/2)csc^2(x/2) ?

The integral of cot^3(x/2)csc^2(x/2) is -1/2 cot^4(x/2)+C

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integral of cot^3(x/2)csc^2(x/2)

\int cot^{3} (\frac{x}{2}) csc^{2} (\frac{x}{2}) d x

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

Solution steps

\int cot^{3} (\frac{x}{2}) csc^{2} (\frac{x}{2}) d x

Apply u-substitution

: \int - 2 u^{3} d u

= \int - 2 u^{3} d u

Take the constant out:

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

= - 2 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - 2 \cdot \frac{u ^{4}}{4}

Substitute back

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

= - 2 \cdot \frac{cot ^{4} ( \frac{x}{2} )}{4}

Simplify

- 2 \cdot \frac{cot ^{4} ( \frac{x}{2} )}{4} : - \frac{1}{2} cot^{4} (\frac{x}{2})

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x ^{2} - 2 x ) ^{3}}{5 x})

\int 8 x^{3} e^{4 x} d x

e^{- 2 t} t^{\frac{3}{2}}

f (x) = - 8 x^{2} - 7 x, at x = - 4

\int \frac{x ^{2} - 3 x + 12}{( x ^{2} - 4 x + 11 ) ^{2}} d x

What is the integral of cot^3(x/2)csc^2(x/2) ?
The integral of cot^3(x/2)csc^2(x/2) is -1/2 cot^4(x/2)+C