What is the integral of \sqrt[5]{tan(6x)}sec^2(6x) ?

The integral of \sqrt[5]{tan(6x)}sec^2(6x) is 5/36 tan^{6/5}(6x)+C

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integral of \sqrt[5]{tan(6x)}sec^2(6x)

\frac{5}{36} tan^{\frac{6}{5}} (6 x) + C

Solution steps

Apply u-substitution

: \int \frac{5 u ^{5}}{6} d u

= \int \frac{5 u ^{5}}{6} d u

Take the constant out:

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

= \frac{5}{6} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{5}{6} \cdot \frac{u ^{6}}{6}

Substitute back

Simplify

= \frac{5}{36} tan^{\frac{6}{5}} (6 x)

Add a constant to the solution

= \frac{5}{36} tan^{\frac{6}{5}} (6 x) + C

\frac{d}{d x} (4 e^{3 x} + 12 x e^{3 x})

sin (5 t)

\frac{\partial}{\partial x} (- 3 x y (- 3 - x - y))

\frac{d y}{d x} = \frac{( x ^{2} + y )}{x ^{3}}

f (x) = ln (x + 2)

What is the integral of \sqrt[5]{tan(6x)}sec^2(6x) ?
The integral of \sqrt[5]{tan(6x)}sec^2(6x) is 5/36 tan^{6/5}(6x)+C