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

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

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integral of 3sec^2(2t)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \int sec^{2} (u) \frac{1}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} tan (u)

Substitute back

u = 2 t

= 3 \cdot \frac{1}{2} tan (2 t)

Simplify

3 \cdot \frac{1}{2} tan (2 t) : \frac{3}{2} tan (2 t)

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

Add a constant to the solution

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

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