What is the integral of 7/9 sec^2(x/9) ?

The integral of 7/9 sec^2(x/9) is 7tan(x/9)+C

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integral of 7/9 sec^2(x/9)

\int \frac{7}{9} sec^{2} (\frac{x}{9}) d x

7 tan (\frac{x}{9}) + C

Solution steps

\int \frac{7}{9} sec^{2} (\frac{x}{9}) d x

Take the constant out:

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

= \frac{7}{9} \cdot \int sec^{2} (\frac{x}{9}) d x

Apply u-substitution

: \int sec^{2} (u) \cdot 9 d u

= \frac{7}{9} \cdot \int sec^{2} (u) \cdot 9 d u

Take the constant out:

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

= \frac{7}{9} \cdot 9 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{7}{9} \cdot 9 tan (u)

Substitute back

u = \frac{x}{9}

= \frac{7}{9} \cdot 9 tan (\frac{x}{9})

Simplify

\frac{7}{9} \cdot 9 tan (\frac{x}{9}) : 7 tan (\frac{x}{9})

= 7 tan (\frac{x}{9})

Add a constant to the solution

= 7 tan (\frac{x}{9}) + C

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What is the integral of 7/9 sec^2(x/9) ?
The integral of 7/9 sec^2(x/9) is 7tan(x/9)+C