What is the integral of θarcsec(9θ) ?

The integral of θarcsec(9θ) is 1/162 (81θ^2arcsec(9θ)-sqrt(81θ^2-1))+C

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integral of θarcsec(9θ)

\int θ arcsec (9 θ) d θ

\frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1) + C

Solution steps

\int θ arcsec (9 θ) d θ

Apply u-substitution

= \int \frac{u arcsec ( u )}{81} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{81} \cdot \int u arcsec (u) d u

Apply Integration By Parts

= \frac{1}{81} (\frac{1}{2} u^{2} arcsec (u) - \int \frac{u ^{2}}{2 u ^{2} u ^{2} - 1} d u)

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

= \frac{1}{81} (\frac{1}{2} u^{2} arcsec (u) - \frac{1}{2} u^{2} - 1)

Substitute back

u = 9 θ

= \frac{1}{81} (\frac{1}{2} (9 θ)^{2} arcsec (9 θ) - \frac{1}{2} (9 θ)^{2} - 1)

Simplify

\frac{1}{81} (\frac{1}{2} (9 θ)^{2} arcsec (9 θ) - \frac{1}{2} (9 θ)^{2} - 1) : \frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1)

= \frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1)

Add a constant to the solution

= \frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1) + C

What is the integral of θarcsec(9θ) ?
The integral of θarcsec(9θ) is 1/162 (81θ^2arcsec(9θ)-sqrt(81θ^2-1))+C