What is the integral of θarcsec(6θ) ?

The integral of θarcsec(6θ) is 1/36 (18θ^2arcsec(6θ)-1/2 sqrt(36θ^2-1))+C

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integral of θarcsec(6θ)

\int θ arcsec (6 θ) d θ

\frac{1}{36} (18 θ^{2} arcsec (6 θ) - \frac{1}{2} 36 θ^{2} - 1) + C

Solution steps

\int θ arcsec (6 θ) d θ

Apply u-substitution

: \int \frac{u arcsec ( u )}{36} d u

= \int \frac{u arcsec ( u )}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u arcsec (u) d u

Apply Integration By Parts

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

= \frac{1}{36} (\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}{36} (\frac{1}{2} u^{2} arcsec (u) - \frac{1}{2} u^{2} - 1)

Substitute back

u = 6 θ

= \frac{1}{36} (\frac{1}{2} (6 θ)^{2} arcsec (6 θ) - \frac{1}{2} (6 θ)^{2} - 1)

Simplify

\frac{1}{36} (\frac{1}{2} (6 θ)^{2} arcsec (6 θ) - \frac{1}{2} (6 θ)^{2} - 1) : \frac{1}{36} (18 θ^{2} arcsec (6 θ) - \frac{1}{2} 36 θ^{2} - 1)

= \frac{1}{36} (18 θ^{2} arcsec (6 θ) - \frac{1}{2} 36 θ^{2} - 1)

Add a constant to the solution

= \frac{1}{36} (18 θ^{2} arcsec (6 θ) - \frac{1}{2} 36 θ^{2} - 1) + C

\int (\frac{1}{2})^{r^{2}} d θ

f (x) = \frac{14}{( 1 + e - x ) , at ( 0 , 7 )}

\frac{d}{d x} (ln (\frac{x ^{\frac{4}{3}} cos ( x )}{7 x + 6}))

y = f (x), at x = 1

\int_{0}^{1} \frac{37}{x ^{5}} d x

What is the integral of θarcsec(6θ) ?
The integral of θarcsec(6θ) is 1/36 (18θ^2arcsec(6θ)-1/2 sqrt(36θ^2-1))+C