What is the integral of (sec^3(1/x)tan^2(1/x))/(x^2) ?

The integral of (sec^3(1/x)tan^2(1/x))/(x^2) is

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integral of (sec^3(1/x)tan^2(1/x))/(x^2)

\int \frac{sec ^{3} ( \frac{1}{x} ) tan ^{2} ( \frac{1}{x} )}{x ^{2}} d x

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

Solution steps

\int \frac{sec ^{3} ( \frac{1}{x} ) tan ^{2} ( \frac{1}{x} )}{x ^{2}} d x

Apply u-substitution

= \int - sec^{3} (u) tan^{2} (u) d u

Take the constant out:

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

= - \int sec^{3} (u) tan^{2} (u) d u

Rewrite using trig identities

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

Expand

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= - \int - sec^{3} (u) + sec^{5} (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 sec^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= - (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = \frac{1}{x}

= - (- (\frac{1}{2} sec (\frac{1}{x}) tan (\frac{1}{x}) + \frac{1}{2} ln tan (\frac{1}{x}) + sec (\frac{1}{x})) + \frac{sec ^{4} ( \frac{1}{x} ) sin ( \frac{1}{x} )}{4} + \frac{3}{4} (\frac{1}{2} sec (\frac{1}{x}) tan (\frac{1}{x}) + \frac{1}{2} ln tan (\frac{1}{x}) + sec (\frac{1}{x})))

Simplify

- (- (\frac{1}{2} sec (\frac{1}{x}) tan (\frac{1}{x}) + \frac{1}{2} ln tan (\frac{1}{x}) + sec (\frac{1}{x})) + \frac{sec ^{4} ( \frac{1}{x} ) sin ( \frac{1}{x} )}{4} + \frac{3}{4} (\frac{1}{2} sec (\frac{1}{x}) tan (\frac{1}{x}) + \frac{1}{2} ln tan (\frac{1}{x}) + sec (\frac{1}{x}))) : \frac{1}{8} sec (\frac{1}{x}) tan (\frac{1}{x}) + \frac{1}{8} ln tan (\frac{1}{x}) + sec (\frac{1}{x}) - \frac{1}{4} sec^{4} (\frac{1}{x}) sin (\frac{1}{x})

= \frac{1}{8} sec (\frac{1}{x}) tan (\frac{1}{x}) + \frac{1}{8} ln tan (\frac{1}{x}) + sec (\frac{1}{x}) - \frac{1}{4} sec^{4} (\frac{1}{x}) sin (\frac{1}{x})

Add a constant to the solution

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

What is the integral of (sec^3(1/x)tan^2(1/x))/(x^2) ?
The integral of (sec^3(1/x)tan^2(1/x))/(x^2) is