What is the integral of xcsc^2(4x) ?

The integral of xcsc^2(4x) is -1/4 xcot(4x)+1/16 ln|sin(4x)|+C

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integral of xcsc^2(4x)

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

- \frac{1}{4} x cot (4 x) + \frac{1}{16} ln ∣ sin (4 x) ∣ + C

Solution steps

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

Apply Integration By Parts

: - \frac{1}{4} x cot (4 x) - \int - \frac{1}{4} cot (4 x) d x

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

\int - \frac{1}{4} cot (4 x) d x = - \frac{1}{16} ln ∣ sin (4 x) ∣

= - \frac{1}{4} x cot (4 x) - (- \frac{1}{16} ln ∣ sin (4 x) ∣)

Simplify

= - \frac{1}{4} x cot (4 x) + \frac{1}{16} ln ∣ sin (4 x) ∣

Add a constant to the solution

= - \frac{1}{4} x cot (4 x) + \frac{1}{16} ln ∣ sin (4 x) ∣ + C

\int_{- 1}^{1} 0.09375 (4 - x^{2}) d x

\int e^{- 17 x} d x

\int \frac{x}{16 - x ^{8}} d x

\frac{d y}{d x} = \frac{2 x + y}{3 + 3 y ^{2} - x}, y (0) = 0

\frac{d}{d x} (ln (sin (x)))

What is the integral of xcsc^2(4x) ?
The integral of xcsc^2(4x) is -1/4 xcot(4x)+1/16 ln|sin(4x)|+C