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

The integral of xtan(2x^2) is -1/4 ln|cos(2x^2)|+C

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integral of xtan(2x^2)

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

- \frac{1}{4} ln cos (2 x^{2}) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{tan ( 2 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 (2 u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{2 v} d v

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} (- \frac{1}{2} ln ∣ v ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

cos^{5} (x)

\int 7 x ln (x) d x

\int (2 + cos (x)) d x

\int \frac{1}{2 - x} d x

\int 7 x e^{- 4 x^{2}} d x

What is the integral of xtan(2x^2) ?
The integral of xtan(2x^2) is -1/4 ln|cos(2x^2)|+C