What is the integral of sec^2(8x)tan^2(8x) ?

The integral of sec^2(8x)tan^2(8x) is 1/24 tan^3(8x)+C

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integral of sec^2(8x)tan^2(8x)

\int sec^{2} (8 x) tan^{2} (8 x) d x

\frac{1}{24} tan^{3} (8 x) + C

Solution steps

\int sec^{2} (8 x) tan^{2} (8 x) d x

Apply u-substitution

= \int sec^{2} (u) tan^{2} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec^{2} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int (1 + tan^{2} (u)) tan^{2} (u) d u

Apply u-substitution

= \frac{1}{8} \cdot \int v^{2} d v

Apply the Power Rule

= \frac{1}{8} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{8} \cdot \frac{tan ^{3} ( 8 x )}{3}

Simplify

\frac{1}{8} \cdot \frac{tan ^{3} ( 8 x )}{3} : \frac{1}{24} tan^{3} (8 x)

= \frac{1}{24} tan^{3} (8 x)

Add a constant to the solution

= \frac{1}{24} tan^{3} (8 x) + C

What is the integral of sec^2(8x)tan^2(8x) ?
The integral of sec^2(8x)tan^2(8x) is 1/24 tan^3(8x)+C