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

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

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integral of (sec^3(x))/(tan^4(x))

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

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

Solution steps

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

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

= \int cot^{4} (x) sec^{3} (x) d x

Rewrite using trig identities

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

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 (x)

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

Simplify

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

= - \frac{1}{3} csc^{3} (x)

Add a constant to the solution

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

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