What is the integral of (sec^2(3sqrt(t)))/(sqrt(t)) ?

The integral of (sec^2(3sqrt(t)))/(sqrt(t)) is 2/3 tan(3sqrt(t))+C

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integral of (sec^2(3sqrt(t)))/(sqrt(t))

\int \frac{sec ^{2} ( 3 t )}{t} d t

\frac{2}{3} tan (3 t) + C

Solution steps

\int \frac{sec ^{2} ( 3 t )}{t} d t

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{2}{3} tan (u)

Substitute back

u = 3 t

= \frac{2}{3} tan (3 t)

Add a constant to the solution

= \frac{2}{3} tan (3 t) + C

What is the integral of (sec^2(3sqrt(t)))/(sqrt(t)) ?
The integral of (sec^2(3sqrt(t)))/(sqrt(t)) is 2/3 tan(3sqrt(t))+C