What is the integral of 4tan(3/2 x) ?

The integral of 4tan(3/2 x) is -8/3 ln|cos((3x)/2)|+C

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integral of 4tan(3/2 x)

\int 4 tan (\frac{3}{2} x) d x

- \frac{8}{3} ln cos (\frac{3 x}{2}) + C

Solution steps

\int 4 tan (\frac{3}{2} x) d x

Take the constant out:

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

= 4 \cdot \int tan (\frac{3}{2} x) d x

Rewrite using trig identities

= 4 \cdot \int \frac{sin ( \frac{3 x}{2} )}{cos ( \frac{3 x}{2} )} d x

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= 4 \cdot \int tan (\frac{3 x}{2}) d x

Apply u-substitution

: \int tan (u) \frac{2}{3} d u

= 4 \cdot \int tan (u) \frac{2}{3} d u

Take the constant out:

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

= 4 \cdot \frac{2}{3} \cdot \int tan (u) d u

Rewrite using trig identities

= 4 \cdot \frac{2}{3} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= 4 \cdot \frac{2}{3} \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

= 4 \cdot \frac{2}{3} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{2}{3} (- ln ∣ v ∣)

Substitute back

= 4 \cdot \frac{2}{3} (- ln cos (\frac{3 x}{2}))

Simplify

4 \cdot \frac{2}{3} (- ln cos (\frac{3 x}{2})) : - \frac{8}{3} ln cos (\frac{3 x}{2})

= - \frac{8}{3} ln cos (\frac{3 x}{2})

Add a constant to the solution

= - \frac{8}{3} ln cos (\frac{3 x}{2}) + C

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What is the integral of 4tan(3/2 x) ?
The integral of 4tan(3/2 x) is -8/3 ln|cos((3x)/2)|+C