What is the integral of \sqrt[5]{tan(4x)}sec^2(4x) ?

The integral of \sqrt[5]{tan(4x)}sec^2(4x) is 5/24 tan^{6/5}(4x)+C

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integral of \sqrt[5]{tan(4x)}sec^2(4x)

\int 5 tan (4 x) sec^{2} (4 x) d x

\frac{5}{24} tan^{\frac{6}{5}} (4 x) + C

Solution steps

\int 5 tan (4 x) sec^{2} (4 x) d x

Apply u-substitution

: \int \frac{5 u ^{5}}{4} d u

= \int \frac{5 u ^{5}}{4} d u

Take the constant out:

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

= \frac{5}{4} \cdot \int u^{5} d u

Apply the Power Rule

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

= \frac{5}{4} \cdot \frac{u ^{6}}{6}

Substitute back

u = 5 tan (4 x)

= \frac{5}{4} \cdot \frac{( 5 tan ( 4 x ) ) ^{6}}{6}

Simplify

\frac{5}{4} \cdot \frac{( 5 tan ( 4 x ) ) ^{6}}{6} : \frac{5}{24} tan^{\frac{6}{5}} (4 x)

= \frac{5}{24} tan^{\frac{6}{5}} (4 x)

Add a constant to the solution

= \frac{5}{24} tan^{\frac{6}{5}} (4 x) + C

\int (5 e^{x} - \frac{1}{x}) d x

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

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

\int 12 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

What is the integral of \sqrt[5]{tan(4x)}sec^2(4x) ?
The integral of \sqrt[5]{tan(4x)}sec^2(4x) is 5/24 tan^{6/5}(4x)+C