What is the integral of arcsec(y) ?

The integral of arcsec(y) is yarcsec(y)-ln|y+sqrt(y^2-1)|+C

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integral of arcsec(y)

\int arcsec (y) d y

y arcsec (y) - ln y + y^{2} - 1 + C

Solution steps

\int arcsec (y) d y

Apply Integration By Parts

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

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

\int \frac{y}{y ^{2} y ^{2} - 1} d y = ln y + y^{2} - 1

= y arcsec (y) - ln y + y^{2} - 1

Add a constant to the solution

= y arcsec (y) - ln y + y^{2} - 1 + C

\int sec^{2} (θ) tan^{6} (θ) d θ

e^{p} (p + p p)

\frac{\partial}{\partial x} (\frac{y}{1 + x ^{2} y ^{2}})

y^{^{'}} - a \cdot y + b = 0

y^{^{'}} = \frac{( e ^{(\frac{x}{3})} + y )}{2}

What is the integral of arcsec(y) ?
The integral of arcsec(y) is yarcsec(y)-ln|y+sqrt(y^2-1)|+C