What is the integral of 1/(tan^3(θ)) ?

The integral of 1/(tan^3(θ)) is -1/2 csc^2(θ)-ln|sin(θ)|+C

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integral of 1/(tan^3(θ))

\int \frac{1}{tan ^{3} ( θ )} d θ

- \frac{1}{2} csc^{2} (θ) - ln ∣ sin (θ) ∣ + C

Solution steps

\int \frac{1}{tan ^{3} ( θ )} d θ

Rewrite using trig identities

= \int \frac{( 1 - sin ^{2} ( θ ) ) cos ( θ )}{sin ^{3} ( θ )} d θ

Apply u-substitution

: \int \frac{1 - u ^{2}}{u ^{3}} d u

= \int \frac{1 - u ^{2}}{u ^{3}} d u

Expand

\frac{1 - u ^{2}}{u ^{3}} : \frac{1}{u ^{3}} - \frac{1}{u}

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

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

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{2 u ^{2}} - ln ∣ u ∣

Substitute back

u = sin (θ)

= - \frac{1}{2 sin ^{2} ( θ )} - ln ∣ sin (θ) ∣

Simplify

- \frac{1}{2 sin ^{2} ( θ )} - ln ∣ sin (θ) ∣ : - \frac{1}{2} csc^{2} (θ) - ln ∣ sin (θ) ∣

= - \frac{1}{2} csc^{2} (θ) - ln ∣ sin (θ) ∣

Add a constant to the solution

= - \frac{1}{2} csc^{2} (θ) - ln ∣ sin (θ) ∣ + C

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What is the integral of 1/(tan^3(θ)) ?
The integral of 1/(tan^3(θ)) is -1/2 csc^2(θ)-ln|sin(θ)|+C