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

The integral of csc^5(2x) is 1/2 (-1/4 csc^4(2x)cos(2x)+3/4 (-1/2 cot(2x)csc(2x)+1/2 ln|tan(x)|))+C

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integral of csc^5(2x)

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

\frac{1}{2} (- \frac{1}{4} csc^{4} (2 x) cos (2 x) + \frac{3}{4} (- \frac{1}{2} cot (2 x) csc (2 x) + \frac{1}{2} ln ∣ tan (x) ∣)) + C

Solution steps

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

Apply u-substitution

: \int csc^{5} (u) \frac{1}{2} d u

= \int csc^{5} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int csc^{5} (u) d u

Apply Integral Reduction

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

\int csc^{3} (u) d u = - \frac{1}{2} cot (u) csc (u) + \frac{1}{2} ln tan (\frac{u}{2})

= \frac{1}{2} (- \frac{cos ( u ) csc ^{4} ( u )}{4} + \frac{3}{4} (- \frac{1}{2} cot (u) csc (u) + \frac{1}{2} ln tan (\frac{u}{2})))

Substitute back

u = 2 x

= \frac{1}{2} (- \frac{cos ( 2 x ) csc ^{4} ( 2 x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (2 x) csc (2 x) + \frac{1}{2} ln tan (\frac{2 x}{2})))

Simplify

\frac{1}{2} (- \frac{cos ( 2 x ) csc ^{4} ( 2 x )}{4} + \frac{3}{4} (- \frac{1}{2} cot (2 x) csc (2 x) + \frac{1}{2} ln tan (\frac{2 x}{2}))) : \frac{1}{2} (- \frac{1}{4} csc^{4} (2 x) cos (2 x) + \frac{3}{4} (- \frac{1}{2} cot (2 x) csc (2 x) + \frac{1}{2} ln ∣ tan (x) ∣))

= \frac{1}{2} (- \frac{1}{4} csc^{4} (2 x) cos (2 x) + \frac{3}{4} (- \frac{1}{2} cot (2 x) csc (2 x) + \frac{1}{2} ln ∣ tan (x) ∣))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4} csc^{4} (2 x) cos (2 x) + \frac{3}{4} (- \frac{1}{2} cot (2 x) csc (2 x) + \frac{1}{2} ln ∣ tan (x) ∣)) + C

\frac{\partial}{\partial y} (x ln (t))

\int e^{x t} d x

x \to 10 lim ((3 - \frac{4}{5} x) + 5)

\frac{d}{d x} ((arcsin (2 x))^{2})

e^{- 3 t}, at t = 0

What is the integral of csc^5(2x) ?
The integral of csc^5(2x) is 1/2 (-1/4 csc^4(2x)cos(2x)+3/4 (-1/2 cot(2x)csc(2x)+1/2 ln|tan(x)|))+C