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)

\int 3 tan (4 x) sec^{2} (4 x) d x

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

Solution steps

\int 3 tan (4 x) sec^{2} (4 x) d x

Apply u-substitution

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

= \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{u ^{4}}{4}

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

Substitute back

u = 3 tan (4 x)

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

Simplify

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

= \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

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x \to 0 lim (\frac{sin ( x ^{2} )}{x})

\frac{d}{d x} (20 + 0.6 x)

\frac{e ^{3 x}}{x ^{6}}

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{9}})

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