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

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

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integral of 2xtan(x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} (- ln ∣ v ∣)

Substitute back

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

Simplify

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

= - ln cos (x^{2})

Add a constant to the solution

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

x \to - 3 lim (\frac{4 x}{x - 3})

\int sin (9 x) cos (2 x) d x

\int 15 cot^{4} (x) d x

\frac{\partial}{\partial x} (- 2 y cos^{2} (x))

\frac{d y}{d x} + y = x^{2} + 1

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