What is the integral of sec(pix)tan(pix) ?

The integral of sec(pix)tan(pix) is 1/pi sec(pix)+C

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integral of sec(pix)tan(pix)

\int sec (π x) tan (π x) d x

\frac{1}{π} sec (π x) + C

Solution steps

\int sec (π x) tan (π x) d x

Apply u-substitution

: \int sec (u) tan (u) \frac{1}{π} d u

= \int sec (u) tan (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 (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{π} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{π} \cdot 1 \cdot v

Substitute back

= \frac{1}{π} \cdot 1 \cdot sec (π x)

Simplify

= \frac{1}{π} sec (π x)

Add a constant to the solution

= \frac{1}{π} sec (π x) + C

\int 8 e^{8 x - 3} d x

x \to 6 lim ((\frac{1}{x - 6})^{2})

\frac{d}{d x} (x 3)

y = \frac{25 - x ^{2}}{4 x}

x y^{2} + 3 y^{2} - x^{2} \cdot \frac{d y}{d x} = 0

What is the integral of sec(pix)tan(pix) ?
The integral of sec(pix)tan(pix) is 1/pi sec(pix)+C