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

The integral of \sqrt[5]{tan(2x)}sec^2(2x) is 5/12 tan^{6/5}(2x)+C

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

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

\frac{5}{12} tan^{\frac{6}{5}} (2 x) + C

Solution steps

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

Apply u-substitution

= \int \frac{5 u ^{5}}{2} d u

Take the constant out:

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

= \frac{5}{2} \cdot \int u^{5} d u

Apply the Power Rule

= \frac{5}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = 5 tan (2 x)

= \frac{5}{2} \cdot \frac{( 5 tan ( 2 x ) ) ^{6}}{6}

Simplify

\frac{5}{2} \cdot \frac{( 5 tan ( 2 x ) ) ^{6}}{6} : \frac{5}{12} tan^{\frac{6}{5}} (2 x)

= \frac{5}{12} tan^{\frac{6}{5}} (2 x)

Add a constant to the solution

= \frac{5}{12} tan^{\frac{6}{5}} (2 x) + C

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What is the integral of \sqrt[5]{tan(2x)}sec^2(2x) ?
The integral of \sqrt[5]{tan(2x)}sec^2(2x) is 5/12 tan^{6/5}(2x)+C