What is the integral of 6csc^4(θ) ?

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

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integral of 6csc^4(θ)

\int 6 csc^{4} (θ) d θ

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

Solution steps

\int 6 csc^{4} (θ) d θ

Take the constant out:

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

= 6 \cdot \int csc^{4} (θ) d θ

Apply Integral Reduction

= 6 (- \frac{cos ( θ ) csc ^{3} ( θ )}{3} + \frac{2}{3} \cdot \int csc^{2} (θ) d θ)

\int csc^{2} (θ) d θ = - cot (θ)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{3 - 5 sin ( x )} d x

(1 + x)^{- 1}

\int \frac{2 x}{x ^{2} - 16} + \frac{1}{x} + 2 d x

y^{^{'}} = x

f (x) = (- 2 x^{2} + 4)^{4}

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