What is the integral of zarcsec(z) ?

The integral of zarcsec(z) is 1/2 z^2arcsec(z)-1/2 sqrt(z^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 zarcsec(z)

\int z arcsec (z) d z

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

Solution steps

\int z arcsec (z) d z

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\int e^{t + l n (t)} d t

\int \frac{1}{3 x ^{2} + 7} d x

\int (x^{2} - 2 x) x^{- 3} d x

\int + 2 x e^{y} d x

\int \frac{7}{9} sin (3 x + 6) d x

What is the integral of zarcsec(z) ?
The integral of zarcsec(z) is 1/2 z^2arcsec(z)-1/2 sqrt(z^2-1)+C