What is the integral of sec^2(2x)tan^5(2x) ?

The integral of sec^2(2x)tan^5(2x) is 1/12 tan^6(2x)+C

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integral of sec^2(2x)tan^5(2x)

\int sec^{2} (2 x) tan^{5} (2 x) d x

\frac{1}{12} tan^{6} (2 x) + C

Solution steps

\int sec^{2} (2 x) tan^{5} (2 x) d x

Apply u-substitution

: \int \frac{5 u}{10} d u

= \int \frac{5 u}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int 5 u d u

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{1}{10} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = tan^{5} (2 x)

= \frac{1}{10} \cdot \frac{5}{6} (tan^{5} (2 x))^{\frac{6}{5}}

Simplify

\frac{1}{10} \cdot \frac{5}{6} (tan^{5} (2 x))^{\frac{6}{5}} : \frac{1}{12} tan^{6} (2 x)

= \frac{1}{12} tan^{6} (2 x)

Add a constant to the solution

= \frac{1}{12} tan^{6} (2 x) + C

y = 4 x^{2} + 3 x

1 - x^{2}, x^{2} - 3, x + 3

\frac{\partial}{\partial x} (2 (ln (x)) + y)

\frac{\partial}{\partial x} (x (x - 2))

\int \frac{1}{x ( x + 1 )} d x

What is the integral of sec^2(2x)tan^5(2x) ?
The integral of sec^2(2x)tan^5(2x) is 1/12 tan^6(2x)+C