What is the integral of cot^5(x/2)csc^2(x/2) ?

The integral of cot^5(x/2)csc^2(x/2) is -1/3 cot^6(x/2)+C

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integral of cot^5(x/2)csc^2(x/2)

\int cot^{5} (\frac{x}{2}) csc^{2} (\frac{x}{2}) d x

- \frac{1}{3} cot^{6} (\frac{x}{2}) + C

Solution steps

\int cot^{5} (\frac{x}{2}) csc^{2} (\frac{x}{2}) d x

Apply u-substitution

: \int - 2 u^{5} d u

= \int - 2 u^{5} d u

Take the constant out:

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

= - 2 \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= - 2 \cdot \frac{u ^{6}}{6}

Substitute back

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

= - 2 \cdot \frac{cot ^{6} ( \frac{x}{2} )}{6}

Simplify

- 2 \cdot \frac{cot ^{6} ( \frac{x}{2} )}{6} : - \frac{1}{3} cot^{6} (\frac{x}{2})

= - \frac{1}{3} cot^{6} (\frac{x}{2})

Add a constant to the solution

= - \frac{1}{3} cot^{6} (\frac{x}{2}) + C

sin^{2} (9 x)

y = 28 - x^{2}, y = 0.6 x, y = 3 x

\int \frac{3}{sec ( x ) - 1} d x

x \to a lim (2 f (x))

\frac{d}{d x} (cos (x + 1))

What is the integral of cot^5(x/2)csc^2(x/2) ?
The integral of cot^5(x/2)csc^2(x/2) is -1/3 cot^6(x/2)+C