What is the integral of 1/(sec(ax)+1) ?

The integral of 1/(sec(ax)+1) is 1/a (2arctan(tan((ax)/2))-tan((ax)/2))+C

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integral of 1/(sec(ax)+1)

\int \frac{1}{sec ( a x ) + 1} d x

\frac{1}{a} (2 arctan (tan (\frac{a x}{2})) - tan (\frac{a x}{2})) + C

Solution steps

\int \frac{1}{sec ( a x ) + 1} d x

Apply u-substitution

: \int \frac{1}{a ( sec ( u ) + 1 )} d u

= \int \frac{1}{a ( sec ( u ) + 1 )} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{sec ( u ) + 1} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{1 + v ^{2}} d v

= \frac{1}{a} \cdot \int \frac{1 - v ^{2}}{1 + v ^{2}} d v

\frac{1 - v ^{2}}{1 + v ^{2}} = \frac{2}{1 + v ^{2}} - 1

= \frac{1}{a} \cdot \int \frac{2}{1 + v ^{2}} - 1 d v

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}{a} (\int \frac{2}{1 + v ^{2}} d v - \int 1 d v)

\int \frac{2}{1 + v ^{2}} d v = 2 arctan (v)

\int 1 d v = v

= \frac{1}{a} (2 arctan (v) - v)

Substitute back

= \frac{1}{a} (2 arctan (tan (\frac{a x}{2})) - tan (\frac{a x}{2}))

Add a constant to the solution

= \frac{1}{a} (2 arctan (tan (\frac{a x}{2})) - tan (\frac{a x}{2})) + C

\frac{d y}{d x} = \frac{( ( 4 x ) sec ( \frac{y}{x} ) + y )}{x}

y^{^{'}} = \frac{( y ^{2} )}{x ^{3}}

\frac{\partial}{\partial x} (3 x^{2} y^{2} + 1)

\int \frac{1}{10 x - 3} d x

y = (3 x - 1) (- 3 x^{2} - 4)^{- 3}

What is the integral of 1/(sec(ax)+1) ?
The integral of 1/(sec(ax)+1) is 1/a (2arctan(tan((ax)/2))-tan((ax)/2))+C