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
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.
