What is the integral of 1/(1+sec(3x)) ?

The integral of 1/(1+sec(3x)) is 1/3 (2arctan(tan((3x)/2))-tan((3x)/2))+C

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integral of 1/(1+sec(3x))

\int \frac{1}{1 + sec ( 3 x )} d x

\frac{1}{3} (2 arctan (tan (\frac{3 x}{2})) - tan (\frac{3 x}{2})) + C

Solution steps

\int \frac{1}{1 + sec ( 3 x )} d x

Apply u-substitution

: \int \frac{1}{3 ( 1 + sec ( u ) )} d u

= \int \frac{1}{3 ( 1 + sec ( u ) )} d u

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{3} \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}{3} \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}{3} (\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}{3} (2 arctan (v) - v)

Substitute back

= \frac{1}{3} (2 arctan (tan (\frac{3 x}{2})) - tan (\frac{3 x}{2}))

Add a constant to the solution

= \frac{1}{3} (2 arctan (tan (\frac{3 x}{2})) - tan (\frac{3 x}{2})) + C

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What is the integral of 1/(1+sec(3x)) ?
The integral of 1/(1+sec(3x)) is 1/3 (2arctan(tan((3x)/2))-tan((3x)/2))+C