What is the integral of 3y^2 1/(2y)tan(2yx) ?

The integral of 3y^2 1/(2y)tan(2yx) is -3/4 ln|cos(2yx)|+C

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integral of 3y^2 1/(2y)tan(2yx)

\int 3 y^{2} \frac{1}{2 y} tan (2 y x) d x

- \frac{3}{4} ln ∣ cos (2 y x) ∣ + C

Solution steps

\int 3 y^{2} \frac{1}{2 y} tan (2 y x) d x

Take the constant out:

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

= 3 \cdot \frac{1}{2 y} y^{2} \cdot \int tan (2 y x) d x

Simplify

3 \cdot \frac{1}{2 y} y^{2} : \frac{3 y}{2}

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{3 y}{2} \cdot \frac{1}{2 y} \cdot \int tan (u) d u

Rewrite using trig identities

= \frac{3 y}{2} \cdot \frac{1}{2 y} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= \frac{3 y}{2} \cdot \frac{1}{2 y} \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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{3 y}{2} \cdot \frac{1}{2 y} (- ln ∣ v ∣)

Substitute back

= \frac{3 y}{2} \cdot \frac{1}{2 y} (- ln ∣ cos (2 y x) ∣)

Simplify

\frac{3 y}{2} \cdot \frac{1}{2 y} (- ln ∣ cos (2 y x) ∣) : - \frac{3}{4} ln ∣ cos (2 y x) ∣

= - \frac{3}{4} ln ∣ cos (2 y x) ∣

Add a constant to the solution

= - \frac{3}{4} ln ∣ cos (2 y x) ∣ + C

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What is the integral of 3y^2 1/(2y)tan(2yx) ?
The integral of 3y^2 1/(2y)tan(2yx) is -3/4 ln|cos(2yx)|+C