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

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

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integral of tan^2(3/4 x)

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

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

Solution steps

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

Rewrite using trig identities

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

Simplify

\frac{sin ^{2} ( \frac{3 x}{4} )}{cos ^{2} ( \frac{3 x}{4} )} : tan^{2} (\frac{3 x}{4})

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

Apply u-substitution

: \int tan^{2} (u) \frac{4}{3} d u

= \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

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

Rewrite using trig identities

= \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

= \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)

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

Substitute back

u = \frac{3 x}{4}

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

Simplify

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

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

Add a constant to the solution

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

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