What is the integral of sec^2(4x)tan^3(4x) ?

The integral of sec^2(4x)tan^3(4x) is 1/16 tan^4(4x)+C

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integral of sec^2(4x)tan^3(4x)

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

\frac{1}{16} tan^{4} (4 x) + C

Solution steps

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

Apply u-substitution

: \int sec^{2} (u) tan^{3} (u) \frac{1}{4} d u

= \int sec^{2} (u) tan^{3} (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) tan^{3} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int (1 + tan^{2} (u)) tan^{3} (u) d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{4} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{4} \cdot \frac{v ^{4}}{4}

Substitute back

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

Simplify

\frac{1}{4} \cdot \frac{tan ^{4} ( 4 x )}{4} : \frac{1}{16} tan^{4} (4 x)

= \frac{1}{16} tan^{4} (4 x)

Add a constant to the solution

= \frac{1}{16} tan^{4} (4 x) + C

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What is the integral of sec^2(4x)tan^3(4x) ?
The integral of sec^2(4x)tan^3(4x) is 1/16 tan^4(4x)+C