What is the integral of-729cot^3(θ)csc(θ) ?

The integral of-729cot^3(θ)csc(θ) is -729(-1/3 csc^3(θ)+csc(θ))+C

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integral of-729cot^3(θ)csc(θ)

\int - 729 cot^{3} (θ) csc (θ) d θ

- 729 (- \frac{1}{3} csc^{3} (θ) + csc (θ)) + C

Solution steps

\int - 729 cot^{3} (θ) csc (θ) d θ

Take the constant out:

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

= - 729 \cdot \int cot^{3} (θ) csc (θ) d θ

Rewrite using trig identities

= - 729 \cdot \int \frac{( 1 - sin ^{2} ( θ ) ) cos ( θ )}{sin ^{4} ( θ )} d θ

Apply u-substitution

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

= - 729 \cdot \int \frac{1 - u ^{2}}{u ^{4}} d u

Expand

\frac{1 - u ^{2}}{u ^{4}} : \frac{1}{u ^{4}} - \frac{1}{u ^{2}}

= - 729 \cdot \int \frac{1}{u ^{4}} - \frac{1}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 729 (\int \frac{1}{u ^{4}} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= - 729 (- \frac{1}{3 u ^{3}} - (- \frac{1}{u}))

Substitute back

u = sin (θ)

= - 729 (- \frac{1}{3 sin ^{3} ( θ )} - (- \frac{1}{sin ( θ )}))

Simplify

- 729 (- \frac{1}{3 sin ^{3} ( θ )} - (- \frac{1}{sin ( θ )})) : - 729 (- \frac{1}{3} csc^{3} (θ) + csc (θ))

= - 729 (- \frac{1}{3} csc^{3} (θ) + csc (θ))

Add a constant to the solution

= - 729 (- \frac{1}{3} csc^{3} (θ) + csc (θ)) + C

\int 1 - x^{2} - y^{2} d y

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\frac{d}{d x} (8 x (x^{5} + 1)^{3})

\frac{\partial}{\partial x} (4 + x + x^{2} y)

x \to 0 lim (\frac{4}{x})

What is the integral of-729cot^3(θ)csc(θ) ?
The integral of-729cot^3(θ)csc(θ) is -729(-1/3 csc^3(θ)+csc(θ))+C