Question

How do you evaluate `\int _ { \frac { \pi } { 3} } ^ { \frac { \pi } { 3} } \frac { d x } { 1- \sin x }`?

Answer 1

Recall that `int_a^bf(x)dx` is the area under the graph of `f` from `x=a` to `x=b`.

So, when we have `int_a^af(x)dx`, this is the "area" under `f` from `x=a` to `x=a`, which isn't anything at all: it's an impossibly thin slice of the function, so thin, that it's area doesn't exist such that `int_a^af(x)dx=0`.

We could also justify this with the Fundamental Theorem of Calculus, which says that `int_a^bf'(x)dx=f(b)-f(a)`. When the bounds are the same, we see that `int_a^af'(x)dx=f(a)-f(a)=0`.

So, without having to find the antiderivative of `1/(1-sinx)`, we see that:

`int_(pi/3)^(pi/3)dx/(1-sinx)=0`