What is the integral of \sqrt[5]{tan(8x)}sec^2(8x) ?

The integral of \sqrt[5]{tan(8x)}sec^2(8x) is 5/48 tan^{6/5}(8x)+C

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integral of \sqrt[5]{tan(8x)}sec^2(8x)

\frac{5}{48} tan^{\frac{6}{5}} (8 x) + C

Solution steps

Apply u-substitution

: \int \frac{5 u ^{5}}{8} d u

= \int \frac{5 u ^{5}}{8} d u

Take the constant out:

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

= \frac{5}{8} \cdot \int u^{5} d u

Apply the Power Rule

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

= \frac{5}{8} \cdot \frac{u ^{6}}{6}

Substitute back

Simplify

= \frac{5}{48} tan^{\frac{6}{5}} (8 x)

Add a constant to the solution

= \frac{5}{48} tan^{\frac{6}{5}} (8 x) + C

x \to 5 lim (\frac{\frac{5}{x} - 1}{x - 5})

\frac{\partial}{\partial y} (6 x sin (y - z))

x^{3} cos (2 x)

\frac{d}{d x} (e^{2 x} \cdot sin (3 x))

\frac{1}{e ^{\frac{1}{x} + 1} - 1}

What is the integral of \sqrt[5]{tan(8x)}sec^2(8x) ?
The integral of \sqrt[5]{tan(8x)}sec^2(8x) is 5/48 tan^{6/5}(8x)+C