What is the integral from 0 to 1 of 1/(sin(x)) ?

The integral from 0 to 1 of 1/(sin(x)) is diverges

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integral from 0 to 1 of 1/(sin(x))

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{t a n (\frac{1}{2})} \frac{1}{u} d u

= \int_{0}^{t a n (\frac{1}{2})} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{t a n (\frac{1}{2})}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to 1 of 1/(sin(x)) ?
The integral from 0 to 1 of 1/(sin(x)) is diverges