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

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

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integral of 1/(tan^3(x))

\int \frac{1}{tan ^{3} ( x )} d x

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

Solution steps

\int \frac{1}{tan ^{3} ( x )} d x

Rewrite using trig identities

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

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 (x)

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

Simplify

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

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

Add a constant to the solution

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

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