What is the integral of tan(9x)sec^{5/4}(9x) ?

The integral of tan(9x)sec^{5/4}(9x) is 4/45 sec^{5/4}(9x)+C

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integral of tan(9x)sec^{5/4}(9x)

\int tan (9 x) sec^{\frac{5}{4}} (9 x) d x

\frac{4}{45} sec^{\frac{5}{4}} (9 x) + C

Solution steps

\int tan (9 x) sec^{\frac{5}{4}} (9 x) d x

Apply u-substitution

= \int tan (u) sec^{\frac{5}{4}} (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 tan (u) sec^{\frac{5}{4}} (u) d u

Apply u-substitution

= \frac{1}{9} \cdot \int v^{\frac{1}{4}} d v

Apply the Power Rule

= \frac{1}{9} \cdot \frac{4}{5} v^{\frac{5}{4}}

Substitute back

= \frac{1}{9} \cdot \frac{4}{5} sec^{\frac{5}{4}} (9 x)

Simplify

\frac{1}{9} \cdot \frac{4}{5} sec^{\frac{5}{4}} (9 x) : \frac{4}{45} sec^{\frac{5}{4}} (9 x)

= \frac{4}{45} sec^{\frac{5}{4}} (9 x)

Add a constant to the solution

= \frac{4}{45} sec^{\frac{5}{4}} (9 x) + C

What is the integral of tan(9x)sec^{5/4}(9x) ?
The integral of tan(9x)sec^{5/4}(9x) is 4/45 sec^{5/4}(9x)+C