What is the integral of \sqrt[5]{tan(9x)}sec^2(9x) ?

The integral of \sqrt[5]{tan(9x)}sec^2(9x) is 5/54 tan^{6/5}(9x)+C

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integral of \sqrt[5]{tan(9x)}sec^2(9x)

\frac{5}{54} tan^{\frac{6}{5}} (9 x) + C

Solution steps

Apply u-substitution

: \int \frac{5 u ^{5}}{9} d u

= \int \frac{5 u ^{5}}{9} d u

Take the constant out:

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

= \frac{5}{9} \cdot \int u^{5} d u

Apply the Power Rule

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

= \frac{5}{9} \cdot \frac{u ^{6}}{6}

Substitute back

Simplify

= \frac{5}{54} tan^{\frac{6}{5}} (9 x)

Add a constant to the solution

= \frac{5}{54} tan^{\frac{6}{5}} (9 x) + C

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What is the integral of \sqrt[5]{tan(9x)}sec^2(9x) ?
The integral of \sqrt[5]{tan(9x)}sec^2(9x) is 5/54 tan^{6/5}(9x)+C