What is the integral of sec(4x)tan(4x) ?

The integral of sec(4x)tan(4x) is 1/4 sec(4x)+C

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integral of sec(4x)tan(4x)

\int sec (4 x) tan (4 x) d x

\frac{1}{4} sec (4 x) + C

Solution steps

\int sec (4 x) tan (4 x) d x

Apply u-substitution

: \int \frac{1}{4} d u

= \int \frac{1}{4} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4} u

Substitute back

u = sec (4 x)

= \frac{1}{4} sec (4 x)

Add a constant to the solution

= \frac{1}{4} sec (4 x) + C

y = \frac{1}{x - 3}

\frac{\partial}{\partial x} (x^{2} y^{3} - 1)

f (x) = (x^{4} + 2 x)^{3}, at x = 2

4 e^{x + 2} + e^{- 3}

x y^{^{'}} + y = y^{2}, y (1) = - 15

What is the integral of sec(4x)tan(4x) ?
The integral of sec(4x)tan(4x) is 1/4 sec(4x)+C