What is the integral of \sqrt[3]{tan(4x)}sec^2(4x) ?

The integral of \sqrt[3]{tan(4x)}sec^2(4x) is 3/16 tan^{4/3}(4x)+C

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integral of \sqrt[3]{tan(4x)}sec^2(4x)

\frac{3}{16} tan^{\frac{4}{3}} (4 x) + C

Solution steps

Apply u-substitution

= \int \frac{3 u ^{3}}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int u^{3} d u

Apply the Power Rule

= \frac{3}{4} \cdot \frac{u ^{4}}{4}

Substitute back

Simplify

= \frac{3}{16} tan^{\frac{4}{3}} (4 x)

Add a constant to the solution

= \frac{3}{16} tan^{\frac{4}{3}} (4 x) + C

What is the integral of \sqrt[3]{tan(4x)}sec^2(4x) ?
The integral of \sqrt[3]{tan(4x)}sec^2(4x) is 3/16 tan^{4/3}(4x)+C