What is the integral of csc^4(3-2x) ?

The integral of csc^4(3-2x) is 1/6 csc^3(3-2x)cos(3-2x)+1/3 cot(3-2x)+C

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integral of csc^4(3-2x)

\int csc^{4} (3 - 2 x) d x

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

Solution steps

\int csc^{4} (3 - 2 x) d x

Apply u-substitution

: \int - \frac{1}{2} csc^{4} (u) d u

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

Apply Integral Reduction

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

\int csc^{2} (u) d u = - cot (u)

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

Substitute back

u = 3 - 2 x

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

Simplify

- \frac{1}{2} (- \frac{cos ( 3 - 2 x ) csc ^{3} ( 3 - 2 x )}{3} + \frac{2}{3} (- cot (3 - 2 x))) : \frac{1}{6} csc^{3} (3 - 2 x) cos (3 - 2 x) + \frac{1}{3} cot (3 - 2 x)

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

Add a constant to the solution

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

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What is the integral of csc^4(3-2x) ?
The integral of csc^4(3-2x) is 1/6 csc^3(3-2x)cos(3-2x)+1/3 cot(3-2x)+C