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

The integral of \sqrt[3]{tan(2x)}sec^2(2x) is 3/8 tan^{4/3}(2x)+C

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

\frac{3}{8} tan^{\frac{4}{3}} (2 x) + C

Solution steps

Apply u-substitution

: \int \frac{3 u ^{3}}{2} d u

= \int \frac{3 u ^{3}}{2} d u

Take the constant out:

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

= \frac{3}{2} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{3}{2} \cdot \frac{u ^{4}}{4}

Substitute back

Simplify

= \frac{3}{8} tan^{\frac{4}{3}} (2 x)

Add a constant to the solution

= \frac{3}{8} tan^{\frac{4}{3}} (2 x) + C

x^{2} \frac{d y}{d x} + x y = 3

\frac{d}{d x} ((1 + cos (2 x))^{- 4})

\frac{d}{d x} (\frac{1}{( 3 + x ) ^{2}})

\int e^{6 x + 7} d x

\frac{3}{1 + x ^{2}}

What is the integral of \sqrt[3]{tan(2x)}sec^2(2x) ?
The integral of \sqrt[3]{tan(2x)}sec^2(2x) is 3/8 tan^{4/3}(2x)+C