What is the integral of 8cot(e^{4x})e^{4x} ?

The integral of 8cot(e^{4x})e^{4x} is 2ln|sin(e^{4x})|+C

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integral of 8cot(e^{4x})e^{4x}

\int 8 cot (e^{4 x}) e^{4 x} d x

2 ln sin (e^{4 x}) + C

Solution steps

\int 8 cot (e^{4 x}) e^{4 x} d x

Take the constant out:

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

= 8 \cdot \int cot (e^{4 x}) e^{4 x} d x

Apply u-substitution

: \int cot (e^{u}) e^{u} \frac{1}{4} d u

= 8 \cdot \int cot (e^{u}) e^{u} \frac{1}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int cot (e^{u}) e^{u} d u

Apply u-substitution

: \int cot (v) d v

= 8 \cdot \frac{1}{4} \cdot \int cot (v) d v

Rewrite using trig identities

= 8 \cdot \frac{1}{4} \cdot \int \frac{cos ( v )}{sin ( v )} d v

Apply u-substitution

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= 8 \cdot \frac{1}{4} ln ∣ w ∣

Substitute back

= 8 \cdot \frac{1}{4} ln sin (e^{4 x})

Simplify

8 \cdot \frac{1}{4} ln sin (e^{4 x}) : 2 ln sin (e^{4 x})

= 2 ln sin (e^{4 x})

Add a constant to the solution

= 2 ln sin (e^{4 x}) + C

x \to 3 lim (x - 3)

\frac{d}{d x} ((1 + x)^{- \frac{1}{2}})

y^{^{'}} = - sin (x) - sin (x) y, y (0) = 8

x \to 0 + lim (\frac{1}{e ^{x} - 1})

y^{^{''}} + 16 y = sec^{2} (4 x)

What is the integral of 8cot(e^{4x})e^{4x} ?
The integral of 8cot(e^{4x})e^{4x} is 2ln|sin(e^{4x})|+C