What is the integral of sec^3(pix) ?

The integral of sec^3(pix) is 1/pi ((sec^2(pix)sin(pix))/2+1/2 ln|tan(pix)+sec(pix)|)+C

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integral of sec^3(pix)

\int sec^{3} (π x) d x

\frac{1}{π} (\frac{sec ^{2} ( π x ) sin ( π x )}{2} + \frac{1}{2} ln ∣ tan (π x) + sec (π x) ∣) + C

Solution steps

\int sec^{3} (π x) d x

Apply u-substitution

: \int sec^{3} (u) \frac{1}{π} d u

= \int sec^{3} (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{π} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{π} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = π x

= \frac{1}{π} (\frac{sec ^{2} ( π x ) sin ( π x )}{2} + \frac{1}{2} ln ∣ tan (π x) + sec (π x) ∣)

Add a constant to the solution

= \frac{1}{π} (\frac{sec ^{2} ( π x ) sin ( π x )}{2} + \frac{1}{2} ln ∣ tan (π x) + sec (π x) ∣) + C

\frac{d}{d x} (e^{x e^{y}})

\frac{d}{d x} (- cot (2 x))

\frac{d}{d x} (cos (x^{2}))

\frac{d y}{d x} = \frac{6}{5} (1 - \frac{y}{8})

2 y^{^{'}} = 0

What is the integral of sec^3(pix) ?
The integral of sec^3(pix) is 1/pi ((sec^2(pix)sin(pix))/2+1/2 ln|tan(pix)+sec(pix)|)+C