What is the integral of sec^2(9θ) ?

The integral of sec^2(9θ) is 1/9 tan(9θ)+C

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integral of sec^2(9θ)

\int sec^{2} (9 θ) d θ

\frac{1}{9} tan (9 θ) + C

Solution steps

\int sec^{2} (9 θ) d θ

Apply u-substitution

: \int sec^{2} (u) \frac{1}{9} d u

= \int sec^{2} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{9} tan (u)

Substitute back

u = 9 θ

= \frac{1}{9} tan (9 θ)

Add a constant to the solution

= \frac{1}{9} tan (9 θ) + C

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