What is the integral from 0 to infinity of 9sin^2(x) ?

The integral from 0 to infinity of 9sin^2(x) is diverges

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integral from 0 to infinity of 9sin^2(x)

\int_{0}^{\infty} 9 sin^{2} (x) d x

diverges

Solution steps

\int_{0}^{\infty} 9 sin^{2} (x) d x

Compute the indefinite integral:

\int 9 sin^{2} (x) d x = \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

Compute the boundaries:

\int_{0}^{\infty} 9 sin^{2} (x) d x =

diverges

= diverges

(x - y^{2} x) d x + (y - x^{2} y) d y = 0

x = 6 - y^{2}, x = y

y = x^{3} - 3 x + 2

x^{3} - 3 x + 2

y = 3 x - 2, x = 2, x = 4, y = 0

What is the integral from 0 to infinity of 9sin^2(x) ?
The integral from 0 to infinity of 9sin^2(x) is diverges