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

The integral of (sec(x))/(tan^3(x)) is 1/8 (tan^2(x/2)-4ln|tan(x/2)|-cot^2(x/2))+C

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

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

\frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2})) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{( - u ^{2} + 1 ) ^{2}}{4 u ^{3}} d u

= \int \frac{( - u ^{2} + 1 ) ^{2}}{4 u ^{3}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{( - u ^{2} + 1 ) ^{2}}{u ^{3}} d u

Expand

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

= \frac{1}{4} \cdot \int u - \frac{2}{u} + \frac{1}{u ^{3}} 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

= \frac{1}{4} (\int u d u - \int \frac{2}{u} d u + \int \frac{1}{u ^{3}} d u)

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= \frac{1}{4} (\frac{u ^{2}}{2} - 2 ln ∣ u ∣ - \frac{1}{2 u ^{2}})

Substitute back

u = tan (\frac{x}{2})

= \frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} - 2 ln tan (\frac{x}{2}) - \frac{1}{2 tan ^{2} ( \frac{x}{2} )})

Simplify

\frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} - 2 ln tan (\frac{x}{2}) - \frac{1}{2 tan ^{2} ( \frac{x}{2} )}) : \frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2}))

= \frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2})) + C

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What is the integral of (sec(x))/(tan^3(x)) ?
The integral of (sec(x))/(tan^3(x)) is 1/8 (tan^2(x/2)-4ln|tan(x/2)|-cot^2(x/2))+C