What is the integral of (sec^3(θ))/(tan^4(θ)) ?

The integral of (sec^3(θ))/(tan^4(θ)) is -1/3 csc^3(θ)+C

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integral of (sec^3(θ))/(tan^4(θ))

\int \frac{sec ^{3} ( θ )}{tan ^{4} ( θ )} d θ

- \frac{1}{3} csc^{3} (θ) + C

Solution steps

\int \frac{sec ^{3} ( θ )}{tan ^{4} ( θ )} d θ

Rewrite using trig identities

= \int \frac{cos ( θ )}{sin ^{4} ( θ )} d θ

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = sin (θ)

= - \frac{1}{3 sin ^{3} ( θ )}

Simplify

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

= - \frac{1}{3} csc^{3} (θ)

Add a constant to the solution

= - \frac{1}{3} csc^{3} (θ) + C

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What is the integral of (sec^3(θ))/(tan^4(θ)) ?
The integral of (sec^3(θ))/(tan^4(θ)) is -1/3 csc^3(θ)+C