What is the integral of arcsec(sqrt(t)) ?

The integral of arcsec(sqrt(t)) is tarcsec(sqrt(t))-sqrt(t-1)+C

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integral of arcsec(sqrt(t))

\int arcsec (t) d t

t arcsec (t) - t - 1 + C

Solution steps

\int arcsec (t) d t

Apply Integration By Parts

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

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

\int \frac{1}{2 t - 1} d t = t - 1

= t arcsec (t) - t - 1

Add a constant to the solution

= t arcsec (t) - t - 1 + C

\frac{d}{d x} (\frac{1}{4 - 3 x})

- \frac{x}{y}

\int sin (x) 1 + cos^{2} (x) d x

\frac{d P}{d t} = P (1 0^{- 1} - P \cdot 1 0^{- 7}), P (0) = 7000

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

What is the integral of arcsec(sqrt(t)) ?
The integral of arcsec(sqrt(t)) is tarcsec(sqrt(t))-sqrt(t-1)+C