What is the integral of 4xtan(x^2) ?

The integral of 4xtan(x^2) is -2ln|cos(x^2)|+C

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integral of 4xtan(x^2)

\int 4 x tan (x^{2}) d x

- 2 ln cos (x^{2}) + C

Solution steps

\int 4 x tan (x^{2}) d x

Take the constant out:

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

= 4 \cdot \int x tan (x^{2}) d x

Apply u-substitution

: \int \frac{tan ( u )}{2} d u

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int tan (u) d u

Rewrite using trig identities

= 4 \cdot \frac{1}{2} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= 4 \cdot \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

= 4 \cdot \frac{1}{2} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{1}{2} (- ln ∣ v ∣)

Substitute back

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

Simplify

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

= - 2 ln cos (x^{2})

Add a constant to the solution

= - 2 ln cos (x^{2}) + C

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What is the integral of 4xtan(x^2) ?
The integral of 4xtan(x^2) is -2ln|cos(x^2)|+C