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

The integral of 2/(x^3)tan(1/(x^2)) is ln|cos(1/(x^2))|+C

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

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

ln cos (\frac{1}{x ^{2}}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int - \frac{tan ( u )}{2} d u

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int tan (u) d u)

Rewrite using trig identities

= 2 (- \frac{1}{2} \cdot \int \frac{sin ( u )}{cos ( u )} d u)

Apply u-substitution

= 2 (- \frac{1}{2} \cdot \int - \frac{1}{v} d v)

Take the constant out:

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

= 2 (- \frac{1}{2} (- \int \frac{1}{v} d v))

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- \frac{1}{2} (- ln ∣ v ∣))

Substitute back

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

Simplify

2 (- \frac{1}{2} (- ln cos (\frac{1}{x ^{2}}))) : ln cos (\frac{1}{x ^{2}})

= ln cos (\frac{1}{x ^{2}})

Add a constant to the solution

= ln cos (\frac{1}{x ^{2}}) + C

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