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

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

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integral from-1 to 1 of 1/(sin(x))

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

diverges

Solution steps

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

Take the denominator(s) of

\frac{1}{sin ( x )}

and compare to zero:

x = 0

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_{- 1}^{0} \frac{1}{sin ( x )} d x + \int_{0}^{1} \frac{1}{sin ( x )} d x

\int_{- 1}^{0} \frac{1}{sin ( x )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{2} 1 - (x - 1)^{2} d x

\int_{4}^{8} (- 6 x + 5) d x

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

\int_{0}^{2 π} sin (x) \cdot cos (x) d x

\int_{- \frac{π}{4}}^{0} \frac{1}{cos ^{2} ( x )} d x

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