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

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

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

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

diverges

Solution steps

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

Multiply by the conjugate of

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

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

Take the denominator(s) of

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

and compare to zero:

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 3}^{2} (2 x^{3} - 4 x) d x

\int_{0}^{2} 4 x - x^{2} d x

\int_{1}^{5} t^{2} - 2 t - 8 d t

\int_{- 3}^{2} 12 - x^{2} - x d x

\int_{2}^{3} x^{2} - x - 2 d x

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