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

The integral of tan(4x)sec^2(4x) is 1/8 tan^2(4x)+C

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integral of tan(4x)sec^2(4x)

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

\frac{1}{8} tan^{2} (4 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (4 x)

= \frac{1}{4} \cdot \frac{tan ^{2} ( 4 x )}{2}

Simplify

\frac{1}{4} \cdot \frac{tan ^{2} ( 4 x )}{2} : \frac{1}{8} tan^{2} (4 x)

= \frac{1}{8} tan^{2} (4 x)

Add a constant to the solution

= \frac{1}{8} tan^{2} (4 x) + C

x^{2} - e^{y^{2}}

y = arctan (e^{x})

\int 3 x^{2} - 6 x d x

f (x) = x^{2} + x - 56, g (x) = - x^{2} + 3 x + 4

1 + \frac{a}{r} - \frac{b}{r ^{3}}

What is the integral of tan(4x)sec^2(4x) ?
The integral of tan(4x)sec^2(4x) is 1/8 tan^2(4x)+C