What is the integral of (e^{6x})/(e^{6x)sin^2(4x)} ?

The integral of (e^{6x})/(e^{6x)sin^2(4x)} is -1/4 cot(4x)+C

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integral of (e^{6x})/(e^{6x)sin^2(4x)}

\int \frac{e ^{6 x}}{e ^{6 x} sin ^{2} ( 4 x )} d x

- \frac{1}{4} cot (4 x) + C

Solution steps

\int \frac{e ^{6 x}}{e ^{6 x} sin ^{2} ( 4 x )} d x

Simplify

\frac{e ^{6 x}}{e ^{6 x} sin ^{2} ( 4 x )} : csc^{2} (4 x)

= \int csc^{2} (4 x) d x

Apply u-substitution

: \int csc^{2} (u) \frac{1}{4} d u

= \int csc^{2} (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{4} (- cot (u))

Substitute back

u = 4 x

= \frac{1}{4} (- cot (4 x))

Simplify

= - \frac{1}{4} cot (4 x)

Add a constant to the solution

= - \frac{1}{4} cot (4 x) + C

x \to 2 lim ((\frac{1}{x})^{3})

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

\frac{d}{d x} (\frac{2}{x - 3})

\frac{d}{d x} ((\frac{x ^{2} + 1}{x ^{2} - 1})^{3})

\frac{\partial}{\partial x} (e^{x + 7 y})

What is the integral of (e^{6x})/(e^{6x)sin^2(4x)} ?
The integral of (e^{6x})/(e^{6x)sin^2(4x)} is -1/4 cot(4x)+C