What is the integral of 2/9 sec((2x)/3) ?

The integral of 2/9 sec((2x)/3) is 1/3 ln|tan((2x)/3)+sec((2x)/3)|+C

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integral of 2/9 sec((2x)/3)

\int \frac{2}{9} sec (\frac{2 x}{3}) d x

\frac{1}{3} ln tan (\frac{2 x}{3}) + sec (\frac{2 x}{3}) + C

Solution steps

\int \frac{2}{9} sec (\frac{2 x}{3}) d x

Take the constant out:

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

= \frac{2}{9} \cdot \int sec (\frac{2 x}{3}) d x

Apply u-substitution

: \int sec (u) \frac{3}{2} d u

= \frac{2}{9} \cdot \int sec (u) \frac{3}{2} d u

Take the constant out:

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

= \frac{2}{9} \cdot \frac{3}{2} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{2}{9} \cdot \frac{3}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = \frac{2 x}{3}

= \frac{2}{9} \cdot \frac{3}{2} ln tan (\frac{2 x}{3}) + sec (\frac{2 x}{3})

Simplify

\frac{2}{9} \cdot \frac{3}{2} ln tan (\frac{2 x}{3}) + sec (\frac{2 x}{3}) : \frac{1}{3} ln tan (\frac{2 x}{3}) + sec (\frac{2 x}{3})

= \frac{1}{3} ln tan (\frac{2 x}{3}) + sec (\frac{2 x}{3})

Add a constant to the solution

= \frac{1}{3} ln tan (\frac{2 x}{3}) + sec (\frac{2 x}{3}) + C

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What is the integral of 2/9 sec((2x)/3) ?
The integral of 2/9 sec((2x)/3) is 1/3 ln|tan((2x)/3)+sec((2x)/3)|+C