What is the integral of (sec(mx)+tan(mx))^2 ?

The integral of (sec(mx)+tan(mx))^2 is -x+2/m tan(mx)+2/m sec(mx)+C

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integral of (sec(mx)+tan(mx))^2

\int (sec (m x) + tan (m x))^{2} d x

- x + \frac{2}{m} tan (m x) + \frac{2}{m} sec (m x) + C

Solution steps

\int (sec (m x) + tan (m x))^{2} d x

Apply u-substitution

: \int \frac{( sec ( u ) + tan ( u ) ) ^{2}}{m} d u

= \int \frac{( sec ( u ) + tan ( u ) ) ^{2}}{m} d u

Take the constant out:

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

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

Expand

(sec (u) + tan (u))^{2} : sec^{2} (u) + 2 sec (u) tan (u) + tan^{2} (u)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{m} (\int sec^{2} (u) d u + \int 2 sec (u) tan (u) d u + \int tan^{2} (u) d u)

\int sec^{2} (u) d u = tan (u)

\int 2 sec (u) tan (u) d u = 2 sec (u)

\int tan^{2} (u) d u = - u + tan (u)

= \frac{1}{m} (tan (u) + 2 sec (u) - u + tan (u))

Substitute back

u = m x

= \frac{1}{m} (tan (m x) + 2 sec (m x) - m x + tan (m x))

Simplify

\frac{1}{m} (tan (m x) + 2 sec (m x) - m x + tan (m x)) : - x + \frac{2}{m} tan (m x) + \frac{2}{m} sec (m x)

= - x + \frac{2}{m} tan (m x) + \frac{2}{m} sec (m x)

Add a constant to the solution

= - x + \frac{2}{m} tan (m x) + \frac{2}{m} sec (m x) + C

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What is the integral of (sec(mx)+tan(mx))^2 ?
The integral of (sec(mx)+tan(mx))^2 is -x+2/m tan(mx)+2/m sec(mx)+C