What is the integral from 0 to infinity of 8sin^2(x) ?

The integral from 0 to infinity of 8sin^2(x) is diverges

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integral from 0 to infinity of 8sin^2(x)

\int_{0}^{\infty} 8 sin^{2} (x) d x

diverges

Solution steps

\int_{0}^{\infty} 8 sin^{2} (x) d x

Compute the indefinite integral:

\int 8 sin^{2} (x) d x = 4 (x - \frac{1}{2} sin (2 x)) + C

Compute the boundaries:

\int_{0}^{\infty} 8 sin^{2} (x) d x =

diverges

= diverges

\int arctan (d) x d x

\frac{d}{d θ} (\frac{2}{1 - sin ( θ )})

x \to 5 lim (x^{2} + 5 x + 8)

\frac{x}{1 + 4 x ^{2}}

\frac{d}{d x} (e^{x} sin (x))

What is the integral from 0 to infinity of 8sin^2(x) ?
The integral from 0 to infinity of 8sin^2(x) is diverges