What is the integral of sec^2(8x) ?

The integral of sec^2(8x) is 1/8 tan(8x)+C

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integral of sec^2(8x)

\int sec^{2} (8 x) d x

\frac{1}{8} tan (8 x) + C

Solution steps

\int sec^{2} (8 x) d x

Apply u-substitution

: \int sec^{2} (u) \frac{1}{8} d u

= \int sec^{2} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{8} tan (u)

Substitute back

u = 8 x

= \frac{1}{8} tan (8 x)

Add a constant to the solution

= \frac{1}{8} tan (8 x) + C

(y \cdot e^{- x}) \frac{d y}{d x} = x

\int_{0}^{π} 1 + cos (2 x) d x

\frac{\partial}{\partial x} (x + y^{2} + x y + 3)

y = x^{3} - 3