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

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

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integral of tan^4(x/3)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

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

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{2} \frac{x}{x ^{2} - 1} d x

\frac{∣ x ∣}{5 - x ^{2}}

x^{2} y d x - (x^{3} + y^{3}) d y = 0

f (x) = (2 x^{2} + 3) (5 x - 4)

f (x) = \frac{e ^{x}}{x}

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