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

The integral of tan(2x)sqrt(sec^5(2x)) is 1/5 sec^{5/2}(2x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{5} sec^{\frac{5}{2}} (2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2} tan (u) sec^{\frac{5}{2}} (u) d u

= \int \frac{1}{2} tan (u) sec^{\frac{5}{2}} (u) 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 tan (u) sec^{\frac{5}{2}} (u) d u

Apply u-substitution

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

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

Apply the Power Rule

: \frac{2}{5} v^{\frac{5}{2}}

= \frac{1}{2} \cdot \frac{2}{5} v^{\frac{5}{2}}

Substitute back

= \frac{1}{2} \cdot \frac{2}{5} sec^{\frac{5}{2}} (2 x)

Simplify

\frac{1}{2} \cdot \frac{2}{5} sec^{\frac{5}{2}} (2 x) : \frac{1}{5} sec^{\frac{5}{2}} (2 x)

= \frac{1}{5} sec^{\frac{5}{2}} (2 x)

Add a constant to the solution

= \frac{1}{5} sec^{\frac{5}{2}} (2 x) + C

\frac{d}{d x} ((\frac{4}{5} x)^{4})

\int 4 sec^{2} (2 x) tan (2 x) d x

100 - h^{2}

\frac{d}{d x} ((a - x)^{2})

y^{^{'}} - 2 x y = x

What is the integral of tan(2x)sqrt(sec^5(2x)) ?
The integral of tan(2x)sqrt(sec^5(2x)) is 1/5 sec^{5/2}(2x)+C