What is the integral of xarcsec(2x) ?

The integral of xarcsec(2x) is 1/4 (2x^2arcsec(2x)-1/2 sqrt(4x^2-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xarcsec(2x)

\int x arcsec (2 x) d x

\frac{1}{4} (2 x^{2} arcsec (2 x) - \frac{1}{2} 4 x^{2} - 1) + C

Solution steps

\int x arcsec (2 x) d x

Apply u-substitution

: \int \frac{u arcsec ( u )}{4} d u

= \int \frac{u arcsec ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u arcsec (u) d u

Apply Integration By Parts

: \frac{1}{2} u^{2} arcsec (u) - \int \frac{u ^{2}}{2 u ^{2} u ^{2} - 1} d u

= \frac{1}{4} (\frac{1}{2} u^{2} arcsec (u) - \int \frac{u ^{2}}{2 u ^{2} u ^{2} - 1} d u)

\int \frac{u ^{2}}{2 u ^{2} u ^{2} - 1} d u = \frac{1}{2} u^{2} - 1

= \frac{1}{4} (\frac{1}{2} u^{2} arcsec (u) - \frac{1}{2} u^{2} - 1)

Substitute back

u = 2 x

= \frac{1}{4} (\frac{1}{2} (2 x)^{2} arcsec (2 x) - \frac{1}{2} (2 x)^{2} - 1)

Simplify

\frac{1}{4} (\frac{1}{2} (2 x)^{2} arcsec (2 x) - \frac{1}{2} (2 x)^{2} - 1) : \frac{1}{4} (2 x^{2} arcsec (2 x) - \frac{1}{2} 4 x^{2} - 1)

= \frac{1}{4} (2 x^{2} arcsec (2 x) - \frac{1}{2} 4 x^{2} - 1)

Add a constant to the solution

= \frac{1}{4} (2 x^{2} arcsec (2 x) - \frac{1}{2} 4 x^{2} - 1) + C

\int \frac{1}{e ^{- v} + 1} d v

\int x^{3} \cdot sin (2 x) d x

y = \frac{5}{x ^{4}} - \frac{7}{x}

\frac{3 s}{( s - 1 ) ( s ^{2} + 1 )}

x \to \infty lim (2 x - 2)

What is the integral of xarcsec(2x) ?
The integral of xarcsec(2x) is 1/4 (2x^2arcsec(2x)-1/2 sqrt(4x^2-1))+C