What is the integral from 2 to 4 of arcsec(sqrt(t)) ?

The integral from 2 to 4 of arcsec(sqrt(t)) is -sqrt(3)+(5pi)/6+1

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integral from 2 to 4 of arcsec(sqrt(t))

\int_{2}^{4} arcsec (t) d t

- 3 + \frac{5 π}{6} + 1

Decimal

1.88594 \dots

Solution steps

\int_{2}^{4} arcsec (t) d t

Apply Integration By Parts

: [t arcsec (t) - \int \frac{1}{2 t - 1} d t]_{2}^{4}

= [t arcsec (t) - \int \frac{1}{2 t - 1} d t]_{2}^{4}

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

= [t arcsec (t) - t - 1]_{2}^{4}

Compute the boundaries:

- 3 + \frac{5 π}{6} + 1

= - 3 + \frac{5 π}{6} + 1

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What is the integral from 2 to 4 of arcsec(sqrt(t)) ?
The integral from 2 to 4 of arcsec(sqrt(t)) is -sqrt(3)+(5pi)/6+1