What is the integral of 5xtan(4x^2-1) ?

The integral of 5xtan(4x^2-1) is -5/8 ln|cos(4x^2-1)|+C

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integral of 5xtan(4x^2-1)

\int 5 x tan (4 x^{2} - 1) d x

- \frac{5}{8} ln cos (4 x^{2} - 1) + C

Solution steps

\int 5 x tan (4 x^{2} - 1) d x

Take the constant out:

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

= 5 \cdot \int x tan (4 x^{2} - 1) d x

Apply u-substitution

: \int \frac{tan ( 4 u - 1 )}{2} d u

= 5 \cdot \int \frac{tan ( 4 u - 1 )}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int tan (4 u - 1) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{2} \cdot \int \frac{sin ( 4 u - 1 )}{cos ( 4 u - 1 )} d u

Apply u-substitution

: \int - \frac{1}{4 v} d v

= 5 \cdot \frac{1}{2} \cdot \int - \frac{1}{4 v} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} (- \frac{1}{4} \cdot \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 5 \cdot \frac{1}{2} (- \frac{1}{4} ln ∣ v ∣)

Substitute back

= 5 \cdot \frac{1}{2} (- \frac{1}{4} ln cos (4 x^{2} - 1))

Simplify

5 \cdot \frac{1}{2} (- \frac{1}{4} ln cos (4 x^{2} - 1)) : - \frac{5}{8} ln cos (4 x^{2} - 1)

= - \frac{5}{8} ln cos (4 x^{2} - 1)

Add a constant to the solution

= - \frac{5}{8} ln cos (4 x^{2} - 1) + C

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What is the integral of 5xtan(4x^2-1) ?
The integral of 5xtan(4x^2-1) is -5/8 ln|cos(4x^2-1)|+C