What is the integral of θarcsec(4θ) ?

The integral of θarcsec(4θ) is 1/16 (8θ^2arcsec(4θ)-1/2 sqrt(16θ^2-1))+C

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integral of θarcsec(4θ)

\int θ arcsec (4 θ) d θ

\frac{1}{16} (8 θ^{2} arcsec (4 θ) - \frac{1}{2} 16 θ^{2} - 1) + C

Solution steps

\int θ arcsec (4 θ) d θ

Apply u-substitution

: \int \frac{u arcsec ( u )}{16} d u

= \int \frac{u arcsec ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \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}{16} (\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}{16} (\frac{1}{2} u^{2} arcsec (u) - \frac{1}{2} u^{2} - 1)

Substitute back

u = 4 θ

= \frac{1}{16} (\frac{1}{2} (4 θ)^{2} arcsec (4 θ) - \frac{1}{2} (4 θ)^{2} - 1)

Simplify

\frac{1}{16} (\frac{1}{2} (4 θ)^{2} arcsec (4 θ) - \frac{1}{2} (4 θ)^{2} - 1) : \frac{1}{16} (8 θ^{2} arcsec (4 θ) - \frac{1}{2} 16 θ^{2} - 1)

= \frac{1}{16} (8 θ^{2} arcsec (4 θ) - \frac{1}{2} 16 θ^{2} - 1)

Add a constant to the solution

= \frac{1}{16} (8 θ^{2} arcsec (4 θ) - \frac{1}{2} 16 θ^{2} - 1) + C

y^{^{'}} + 10 y = 2 cos (t)

x y^{^{'}} = 2 y + x^{3} e^{x}

t \to 1 lim (3 t^{3} + 2)

\int \frac{x}{1 + x} d x

\frac{\partial}{\partial x} (- 2 x)

What is the integral of θarcsec(4θ) ?
The integral of θarcsec(4θ) is 1/16 (8θ^2arcsec(4θ)-1/2 sqrt(16θ^2-1))+C