Definite integral problem - sin and its inverse The problem is as follows:
let:
$$
0<a<\frac{\pi}{2} , 0<b<1
$$
show that:
$$
\int_{0}^{a}\sin(x)dx+\int_{0}^{b}\arcsin(x)dx\geq ab
$$
I've tried to calculate the second integral over y and not over x (over $\sin(y)$, didn't got very far.
The solution should be simple, and may involve using double integrals or using the graphs of the sine function and it's inverse in the rectangle that $a$ and $b$ form in $\mathbb{R}^2$.
even an hint would be appreciated.
Thanks
 A: The signs and monotonicty here are such that the integrals measure the area below the curve. 
The parameter ranges allows the decomposition of the rectangle $[0, b] \times [0, \arcsin(b)]$ like this: 
$$
\int\limits_0^b \arcsin(x)\, dx =
\arcsin(b)\, b - \int\limits_0^{\arcsin(b)} \sin(y)\,  dy
$$ 
For $a \le \arcsin(b)$ we have
$$
\int\limits_0^b \arcsin(x)\, dx =
\arcsin(b)\, b 
- \int\limits_0^a \sin(y)\, dy
- \int\limits_a^{\arcsin(b)} \sin(y)\, dy
$$
which means
$$
\int\limits_0^a \sin(x)\,  dx + \int\limits_0^b \arcsin(x)\, dx =
\underbrace{\arcsin(b) \, b}_{\ge ab}
- 
\underbrace{\int\limits_a^{\arcsin(b)} \sin(x)\, dx}_{\ge 0}
\ge a b
$$
We can also decompose $[0, a] \times [0, \sin(a)]$ like this:
$$
\int\limits_0^a \sin(x)\, dx =
a\, \sin(a) - \int\limits_0^{\sin(a)} \arcsin(y)\, dy
$$ 
This gives an estimation for $b \le \sin(a)$:
$$
\int\limits_0^a \sin(x)\, dx + \int\limits_0^b \arcsin(y)\, dy \ge 
a\, b
$$ 
Finally 
$$
b \le \sin(a) \iff \arcsin(b) \le \arcsin(\sin(a))= a
$$
so we cover the other case as well.
