What is the integral of 1/(cos^2(4x)) ?

The integral of 1/(cos^2(4x)) is 1/4 tan(4x)+C

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integral of 1/(cos^2(4x))

\int \frac{1}{cos ^{2} ( 4 x )} d x

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

Solution steps

\int \frac{1}{cos ^{2} ( 4 x )} d x

Rewrite using trig identities

= \int sec^{2} (4 x) d x

Apply u-substitution

: \int sec^{2} (u) \frac{1}{4} d u

= \int sec^{2} (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{4} tan (u)

Substitute back

u = 4 x

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

Add a constant to the solution

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

\frac{d}{d x} (e^{x} - 4)

\int \frac{x ^{2} + 2 x + 5}{( x + 1 ) ( x ^{2} + 1 )} d x

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

\frac{d}{d x} (\frac{1}{5 + x})

\frac{64}{16 + x ^{2}}, 1

What is the integral of 1/(cos^2(4x)) ?
The integral of 1/(cos^2(4x)) is 1/4 tan(4x)+C