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

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

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integral of sec^2(2t)

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

\frac{1}{2} tan (2 t) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 2 t

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

Add a constant to the solution

= \frac{1}{2} tan (2 t) + C

\frac{\partial}{\partial z} (x + yz)

t \to \infty lim (e^{2 t})

\int_{0}^{8} 3 t^{2} - 36 t + 105 d t

x \to 1 lim (g (x))

\int e^{- \frac{1}{2} x} d x