What is the integral of tan(3x)sec^{5/4}(3x) ?

The integral of tan(3x)sec^{5/4}(3x) is 4/15 sec^{5/4}(3x)+C

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integral of tan(3x)sec^{5/4}(3x)

\int tan (3 x) sec^{\frac{5}{4}} (3 x) d x

\frac{4}{15} sec^{\frac{5}{4}} (3 x) + C

Solution steps

\int tan (3 x) sec^{\frac{5}{4}} (3 x) d x

Apply u-substitution

: \int tan (u) sec^{\frac{5}{4}} (u) \frac{1}{3} d u

= \int tan (u) sec^{\frac{5}{4}} (u) \frac{1}{3} 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 tan (u) sec^{\frac{5}{4}} (u) d u

Apply u-substitution

: \int v^{\frac{1}{4}} d v

= \frac{1}{3} \cdot \int v^{\frac{1}{4}} d v

Apply the Power Rule

: \frac{4}{5} v^{\frac{5}{4}}

= \frac{1}{3} \cdot \frac{4}{5} v^{\frac{5}{4}}

Substitute back

= \frac{1}{3} \cdot \frac{4}{5} sec^{\frac{5}{4}} (3 x)

Simplify

\frac{1}{3} \cdot \frac{4}{5} sec^{\frac{5}{4}} (3 x) : \frac{4}{15} sec^{\frac{5}{4}} (3 x)

= \frac{4}{15} sec^{\frac{5}{4}} (3 x)

Add a constant to the solution

= \frac{4}{15} sec^{\frac{5}{4}} (3 x) + C

y = x, (36, 6)

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\frac{d}{d t} (\frac{t}{1 + t})

n = 1 \sum \infty \frac{1}{n e ^{n}}

What is the integral of tan(3x)sec^{5/4}(3x) ?
The integral of tan(3x)sec^{5/4}(3x) is 4/15 sec^{5/4}(3x)+C