What is the integral of (sec^6(x))/(tan^3(x)) ?

The integral of (sec^6(x))/(tan^3(x)) is -1/2 cot^2(x)+2ln|tan(x)|+1/2 tan^2(x)+C

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integral of (sec^6(x))/(tan^3(x))

\int \frac{sec ^{6} ( x )}{tan ^{3} ( x )} d x

- \frac{1}{2} cot^{2} (x) + 2 ln ∣ tan (x) ∣ + \frac{1}{2} tan^{2} (x) + C

Solution steps

\int \frac{sec ^{6} ( x )}{tan ^{3} ( x )} d x

Rewrite using trig identities

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

Apply Trigonometric Substitution

: \int \frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} d u

= \int \frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} d u

Expand

\frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} : \frac{1}{u ^{3}} + \frac{2}{u} + u

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= - \frac{1}{2 u ^{2}} + 2 ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = tan (x)

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

Simplify

- \frac{1}{2 tan ^{2} ( x )} + 2 ln ∣ tan (x) ∣ + \frac{tan ^{2} ( x )}{2} : - \frac{1}{2} cot^{2} (x) + 2 ln ∣ tan (x) ∣ + \frac{1}{2} tan^{2} (x)

= - \frac{1}{2} cot^{2} (x) + 2 ln ∣ tan (x) ∣ + \frac{1}{2} tan^{2} (x)

Add a constant to the solution

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

x \to 0 - lim (e^{\frac{3}{x}})

\frac{\partial}{\partial x} (9)

\frac{d}{d x} ([x + (x + sin^{2} (x))^{3}]^{5})

\int_{0}^{1} 2^{- x} d x

\frac{d}{d x} (\frac{ln ( x )}{x})

What is the integral of (sec^6(x))/(tan^3(x)) ?
The integral of (sec^6(x))/(tan^3(x)) is -1/2 cot^2(x)+2ln|tan(x)|+1/2 tan^2(x)+C