integral from 0 to pi of 1/(1+sin(θ))

解答

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

发散

求解步骤

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

乘以

1 + sin (θ)

的共轭根式

: \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

取

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

的分母，令其等于零:

θ = \frac{π}{2}

如果存在

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

无定义

, \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 =

发散

如果表达式有部分发散，则整个表达式发散

= 发散

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