What is the integral of 6tan^2((3x)/4) ?

The integral of 6tan^2((3x)/4) is 8(-(3x)/4+tan((3x)/4))+C

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integral of 6tan^2((3x)/4)

\int 6 tan^{2} (\frac{3 x}{4}) d x

8 (- \frac{3 x}{4} + tan (\frac{3 x}{4})) + C

Solution steps

\int 6 tan^{2} (\frac{3 x}{4}) d x

Take the constant out:

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

= 6 \cdot \int tan^{2} (\frac{3 x}{4}) d x

Apply u-substitution

= 6 \cdot \int tan^{2} (u) \frac{4}{3} d u

Take the constant out:

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

= 6 \cdot \frac{4}{3} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 6 \cdot \frac{4}{3} \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 \cdot \frac{4}{3} (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= 6 \cdot \frac{4}{3} (- u + tan (u))

Substitute back

u = \frac{3 x}{4}

= 6 \cdot \frac{4}{3} (- \frac{3 x}{4} + tan (\frac{3 x}{4}))

Simplify

6 \cdot \frac{4}{3} (- \frac{3 x}{4} + tan (\frac{3 x}{4})) : 8 (- \frac{3 x}{4} + tan (\frac{3 x}{4}))

= 8 (- \frac{3 x}{4} + tan (\frac{3 x}{4}))

Add a constant to the solution

= 8 (- \frac{3 x}{4} + tan (\frac{3 x}{4})) + C

What is the integral of 6tan^2((3x)/4) ?
The integral of 6tan^2((3x)/4) is 8(-(3x)/4+tan((3x)/4))+C