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

The integral from 2 to 4 of xarcsec(x) is (16arcsec(4)-sqrt(15))/2-(4pi-3sqrt(3))/6

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integral from 2 to 4 of xarcsec(x)

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

\frac{16 arcsec ( 4 ) - 15}{2} - \frac{4 π - 3 3}{6}

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{16 arcsec ( 4 ) - 15}{2} - \frac{4 π - 3 3}{6}

= \frac{16 arcsec ( 4 ) - 15}{2} - \frac{4 π - 3 3}{6}

\int_{\frac{π}{6}}^{\frac{π}{3}} sec^{2} (x) d x

\int_{0}^{1} x e^{\frac{- x ^{2}}{2}} d x

\int_{0}^{5} x^{4} d x

\int_{0}^{\frac{π}{4}} t^{2} sin (4 t) d t

\int_{- 1}^{1} (4 x^{3} - e^{2 x}) d x

What is the integral from 2 to 4 of xarcsec(x) ?
The integral from 2 to 4 of xarcsec(x) is (16arcsec(4)-sqrt(15))/2-(4pi-3sqrt(3))/6