What is the integral of \sqrt[3]{tan(7x)}sec^2(7x) ?

The integral of \sqrt[3]{tan(7x)}sec^2(7x) is 3/28 tan^{4/3}(7x)+C

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integral of \sqrt[3]{tan(7x)}sec^2(7x)

\frac{3}{28} tan^{\frac{4}{3}} (7 x) + C

Solution steps

Apply u-substitution

: \int \frac{3 u ^{3}}{7} d u

= \int \frac{3 u ^{3}}{7} d u

Take the constant out:

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

= \frac{3}{7} \cdot \int u^{3} d u

Apply the Power Rule

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

= \frac{3}{7} \cdot \frac{u ^{4}}{4}

Substitute back

Simplify

= \frac{3}{28} tan^{\frac{4}{3}} (7 x)

Add a constant to the solution

= \frac{3}{28} tan^{\frac{4}{3}} (7 x) + C

\frac{\partial}{\partial y} (y^{2} - 4 x)

\frac{d y}{d t} = y^{\frac{1}{3}} t^{\frac{2}{5}}

y^{^{'}} = 4 + e^{y - 4 x + 1}, y (0) = - 1

38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

\frac{0.5}{( s ^{2} + s + 1 ) \cdot s}

What is the integral of \sqrt[3]{tan(7x)}sec^2(7x) ?
The integral of \sqrt[3]{tan(7x)}sec^2(7x) is 3/28 tan^{4/3}(7x)+C