What is the integral of sec^2(4x) ?

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

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integral of sec^2(4x)

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

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

Solution steps

\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

\int_{- 32}^{0} \frac{1}{4 - x} d x

\frac{d}{d x} (\frac{3 cos ( x )}{3 x ^{3} + 2 x ^{2} - x + 4})

\int 4 cos^{2} (3 x) d x

\int \frac{- sin ( 3 x )}{cos ^{3} ( 3 x )} d x

sin (2 x)