What is the integral of xtan(3x^2) ?

The integral of xtan(3x^2) is -1/6 ln|cos(3x^2)|+C

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integral of xtan(3x^2)

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

- \frac{1}{6} ln cos (3 x^{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{tan ( u )}{6} d u

= \int \frac{tan ( u )}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int tan (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= \frac{1}{6} \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

= \frac{1}{6} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{6} (- ln ∣ v ∣)

Substitute back

= \frac{1}{6} (- ln cos (3 x^{2}))

Simplify

= - \frac{1}{6} ln cos (3 x^{2})

Add a constant to the solution

= - \frac{1}{6} ln cos (3 x^{2}) + C

\int \frac{1}{x ( 3 x + 1 )} d x

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\int cosh (y) d y

\int \frac{x ^{3}}{7 - x ^{4}} d x

\int \frac{2 x + 2}{( x ^{2} + 1 ) ( x - 1 )} d x

What is the integral of xtan(3x^2) ?
The integral of xtan(3x^2) is -1/6 ln|cos(3x^2)|+C