What is the integral of sec^3(2x)tan(2x) ?

The integral of sec^3(2x)tan(2x) is 1/6 sec^3(2x)+C

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integral of sec^3(2x)tan(2x)

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

\frac{1}{6} sec^{3} (2 x) + C

Solution steps

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

Apply u-substitution

: \int sec^{3} (u) tan (u) \frac{1}{2} d u

= \int sec^{3} (u) tan (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{3} (u) tan (u) d u

Apply u-substitution

: \int \frac{1}{3} d v

= \frac{1}{2} \cdot \int \frac{1}{3} d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{3} v

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} sec^{3} (2 x)

Simplify

\frac{1}{2} \cdot \frac{1}{3} sec^{3} (2 x) : \frac{1}{6} sec^{3} (2 x)

= \frac{1}{6} sec^{3} (2 x)

Add a constant to the solution

= \frac{1}{6} sec^{3} (2 x) + C

x^{2} (1 - x)^{2}

\int x^{2} e^{- 2 x^{2}} d x

y = 1 - x^{2}, y = 3 - 3 x

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

\int (arcsin (x)) d x

What is the integral of sec^3(2x)tan(2x) ?
The integral of sec^3(2x)tan(2x) is 1/6 sec^3(2x)+C