What is the integral of 16csc(8x)sec(8x) ?

The integral of 16csc(8x)sec(8x) is ln(tan^2(8x))+C

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integral of 16csc(8x)sec(8x)

\int 16 csc (8 x) sec (8 x) d x

ln (tan^{2} (8 x)) + C

Solution steps

\int 16 csc (8 x) sec (8 x) d x

Take the constant out:

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

= 16 \cdot \int csc (8 x) sec (8 x) d x

Apply u-substitution

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

= 16 \cdot \int csc (u) sec (u) \frac{1}{8} d u

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \int csc (u) sec (u) d u

Apply u-substitution

: \int \frac{v}{v ^{2} - 1} d v

= 16 \cdot \frac{1}{8} \cdot \int \frac{v}{v ^{2} - 1} d v

Apply u-substitution

: \int \frac{1}{2 w} d w

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{2 w} d w

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} ln ∣ w ∣

Substitute back

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} ln sec^{2} (8 x) - 1

Simplify

16 \cdot \frac{1}{8} \cdot \frac{1}{2} ln sec^{2} (8 x) - 1 : ln (tan^{2} (8 x))

= ln (tan^{2} (8 x))

Add a constant to the solution

= ln (tan^{2} (8 x)) + C

x^{2} + 2 x + 1, - 1, 7

y = 8 x, y = x^{2} - 9

\frac{d}{d x} (\frac{400}{1 + 39 e ^{- 0.16 x}})

\int \frac{3 + t + t ^{2}}{t} d t

\int_{5}^{9} \frac{x - 4}{x ^{2}} d x

What is the integral of 16csc(8x)sec(8x) ?
The integral of 16csc(8x)sec(8x) is ln(tan^2(8x))+C