What is the integral from 3 to 4 of 1/(sin(x)) ?

The integral from 3 to 4 of 1/(sin(x)) is diverges

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integral from 3 to 4 of 1/(sin(x))

\int_{3}^{4} \frac{1}{sin ( x )} d x

diverges

Solution steps

\int_{3}^{4} \frac{1}{sin ( x )} d x

Take the denominator(s) of

\frac{1}{sin ( x )}

and compare to zero:

x = π

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{3}^{π} \frac{1}{sin ( x )} d x + \int_{π}^{4} \frac{1}{sin ( x )} d x

\int_{3}^{π} \frac{1}{sin ( x )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

5 cos (2 t) - 3 sin (3 t)

y^{^{''}} - \frac{6}{x} y^{^{'}} + \frac{10}{x ^{2}} y = 0

h \to 0 lim (\frac{h + 5 - 5}{h})

y = \frac{2}{1 - 3 x}, at x = - 1

\int \frac{x + 8}{x ^{2} + 9} d x

What is the integral from 3 to 4 of 1/(sin(x)) ?
The integral from 3 to 4 of 1/(sin(x)) is diverges