What is the integral of sec^3(8x)tan(8x) ?

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

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integral of sec^3(8x)tan(8x)

\int sec^{3} (8 x) tan (8 x) d x

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

Solution steps

\int sec^{3} (8 x) tan (8 x) d x

Apply u-substitution

: \int sec^{3} (u) tan (u) \frac{1}{8} d u

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

Apply u-substitution

: \int \frac{1}{3} d v

= \frac{1}{8} \cdot \int \frac{1}{3} d v

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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