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

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

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

\int_{2}^{4} arcsec (θ) d θ

- 3 + \frac{5 π}{6} + 1

Decimal

1.88594 \dots

Solution steps

\int_{2}^{4} arcsec (θ) d θ

Apply Integration By Parts

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

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

\int \frac{1}{2 θ - 1} d θ = θ - 1

= [θ arcsec (θ) - θ - 1]_{2}^{4}

Compute the boundaries:

- 3 + \frac{5 π}{6} + 1

= - 3 + \frac{5 π}{6} + 1

\int_{0}^{π} (sin (x))^{3} d x

\int_{0}^{1} 2 cos (\frac{π t}{2}) d t

\int_{- 2}^{2} - 4 - x^{2} d x

\int_{s i n (x)}^{c o s (x)} e^{t} d t

\int_{0}^{5} \frac{1}{x - 1} d x

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