What is the integral of xtan(x^2-1) ?

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

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integral of xtan(x^2-1)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int tan (v) d v

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

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{2} \cdot \int - \frac{1}{w} d w

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{2} (- ln ∣ w ∣)

Substitute back

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

Simplify

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

Add a constant to the solution

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

\int x x^{2} + 8 d x

x \to 0 lim (ln (x - 1))

x \frac{d y}{d x} + 2 y = \frac{sin ( x )}{x}

\int cos (\frac{x}{7}) d x

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 2 x - 3

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