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

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

\int 3 tan (7 x) sec^{2} (7 x) d x

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

Solution steps

\int 3 tan (7 x) sec^{2} (7 x) d x

Apply u-substitution

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

Substitute back

u = 3 tan (7 x)

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

Simplify

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

= \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}}

d er i v a t i v e 38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

s l o p e y = \frac{1}{1 - x}, x = 3

\frac{d y}{d x} = 6 y (10 - y)

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