What is the integral from 0 to pi of 1/(1+sin(θ)) ?

The integral from 0 to pi of 1/(1+sin(θ)) is diverges

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integral from 0 to pi of 1/(1+sin(θ))

\int_{0}^{π} \frac{1}{1 + sin ( θ )} d θ

diverges

Solution steps

\int_{0}^{π} \frac{1}{1 + sin ( θ )} d θ

Multiply by the conjugate of

1 + sin (θ) : \frac{cos ^{2} ( θ )}{1 - sin ( θ )}

= \int_{0}^{π} \frac{1}{\frac{c o s ^{2} ( θ )}{1 - s i n ( θ )}} d θ

Take the denominator(s) of

\frac{1}{\frac{c o s ^{2} ( θ )}{1 - s i n ( θ )}}

and compare to zero:

θ = \frac{π}{2}

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_{0}^{\frac{π}{2}} \frac{1}{\frac{c o s ^{2} ( θ )}{1 - s i n ( θ )}} d θ + \int_{\frac{π}{2}}^{π} \frac{1}{\frac{c o s ^{2} ( θ )}{1 - s i n ( θ )}} d θ

\int_{0}^{\frac{π}{2}} \frac{1}{\frac{c o s ^{2} ( θ )}{1 - s i n ( θ )}} d θ =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{0}^{1} t e^{6 t^{2}} d t

\int_{0}^{1} \frac{2 x - 2}{x + 3} d x

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