What is the integral of \sqrt[5]{tan(3x)}sec^2(3x) ?

The integral of \sqrt[5]{tan(3x)}sec^2(3x) is 5/18 tan^{6/5}(3x)+C

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integral of \sqrt[5]{tan(3x)}sec^2(3x)

\frac{5}{18} tan^{\frac{6}{5}} (3 x) + C

Solution steps

Apply u-substitution

: \int \frac{5 u ^{5}}{3} d u

= \int \frac{5 u ^{5}}{3} d u

Take the constant out:

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

= \frac{5}{3} \cdot \int u^{5} d u

Apply the Power Rule

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

= \frac{5}{3} \cdot \frac{u ^{6}}{6}

Substitute back

Simplify

= \frac{5}{18} tan^{\frac{6}{5}} (3 x)

Add a constant to the solution

= \frac{5}{18} tan^{\frac{6}{5}} (3 x) + C

\frac{d}{d x} ((3 x + 7)^{5})

n = 1 \sum \infty \frac{1}{10}

\int \frac{x ^{5}}{2} d x

\int \frac{x \cdot e ^{x}}{( 1 + x ) ^{2}} d x

(8 x)^{4 x}

What is the integral of \sqrt[5]{tan(3x)}sec^2(3x) ?
The integral of \sqrt[5]{tan(3x)}sec^2(3x) is 5/18 tan^{6/5}(3x)+C