What is the integral of 1/3 sec^2(8x) ?

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

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integral of 1/3 sec^2(8x)

\int \frac{1}{3} sec^{2} (8 x) d x

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

Solution steps

\int \frac{1}{3} sec^{2} (8 x) d x

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{2} (8 x) d x

Apply u-substitution

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

= \frac{1}{3} \cdot \int sec^{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}{3} \cdot \frac{1}{8} \cdot \int sec^{2} (u) d u

Use the common integral:

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

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

Substitute back

u = 8 x

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

Simplify

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

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

Add a constant to the solution

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

\int sin (2 t) cos (t) d t

y = (5 x - 4) (2 x^{3} - x^{2} + 1)

y^{^{''}} + 3 y^{^{'}} + 0 = 5

\int_{0}^{\frac{1}{3}} \frac{2}{1 + 9 x ^{2}} d x

f (x) = x^{3} - 16 x + 5, at x = 4

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