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

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

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integral of xtan(1-x^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - tan (v) d v

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Simplify

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

Add a constant to the solution

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

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