What is the integral of 16xtan(4x^2+1) ?

The integral of 16xtan(4x^2+1) is -2ln|cos(4x^2+1)|+C

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integral of 16xtan(4x^2+1)

\int 16 x tan (4 x^{2} + 1) d x

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

Solution steps

\int 16 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

= 16 \cdot \int x tan (4 x^{2} + 1) d x

Apply u-substitution

= 16 \cdot \int \frac{tan ( u )}{8} d u

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \int tan (u) d u

Rewrite using trig identities

= 16 \cdot \frac{1}{8} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

= 16 \cdot \frac{1}{8} \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

= 16 \cdot \frac{1}{8} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 16 \cdot \frac{1}{8} (- ln ∣ v ∣)

Substitute back

= 16 \cdot \frac{1}{8} (- ln cos (4 x^{2} + 1))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 16xtan(4x^2+1) ?
The integral of 16xtan(4x^2+1) is -2ln|cos(4x^2+1)|+C