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

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

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

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

- 3 + \frac{5 π}{6} + 1

Decimal

1.88594 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- 3 + \frac{5 π}{6} + 1

= - 3 + \frac{5 π}{6} + 1

\int_{- 1}^{0} (x^{3} - x) d x

\int_{- 3}^{- 1} x^{2} + x + \frac{1}{4} d x

\int_{0}^{\infty} e^{- 66 x} x d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} - cos (x) d x

\int_{0}^{1} 1 + x d x

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