A function $y=f(x)$ satisfies the condition $f'(x)\sin x+f(x)\cos x=1,f(x)$ being bounded when $x\to 0$.If $I=\int_{0}^{\pi/2}f(x)dx$ A function $y=f(x)$ satisfies the condition $f'(x)\sin x+f(x)\cos x=1,f(x)$ being bounded when $x\to 0$.If $I=\int_{0}^{\pi/2}f(x)dx,$then
$(A)\frac{\pi}{2}<I<\frac{\pi^2}{4}\hspace{1cm}(B)\frac{\pi}{4}<I<\frac{\pi^2}{2}\hspace{1cm}(C)1<I<\frac{\pi}{2}\hspace{1cm}(D)0<I<1$
From the given equation $f'(x)\sin x+f(x)\cos x=1,f(x)$,i could not prove whether $f(x)$ is a monotonically increasing or monotonically decreasing function.Hence i could not find the upper and lower bounds of $I$.Please suggest me some method to solve it.
 A: As a follow-up to G-man's answer, notice that:
$$ I=\int_{0}^{\pi/2}\frac{x}{\sin x}\,dx = 2G, \tag{1}$$
with $G$ being Catalan's constant. $(1)$ follows from the Fourier sine series of the identity function:
$$\forall x\in(-\pi,\pi),\qquad  x = \sum_{n\geq 1}\frac{2(-1)^{n+1}}{n}\,\sin(nx)\tag{2} $$
from which:
$$\begin{eqnarray*} I = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\int_{0}^{\pi/2}\frac{\sin(2nx)}{\sin(x)}\,dx &=&\; 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sum_{j=1}^{n}\frac{(-1)^{j+1}}{2j-1}\\&=&\;4\int_{0}^{1}\frac{\text{arctanh}(x)}{1+x^2}\,dx\\&=&\;2\int_{1}^{+\infty}\frac{\log(y)}{1+y^2}\,dy\\&=&\;2\int_{0}^{1}\frac{-\log x}{1+x^2}\,dx\\&=&\;2\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n-1)^2}\\&=&\,2G.\tag{3}\end{eqnarray*}$$
The last series representation clearly gives $\frac{16}{9}<I<2$, hence both $(A)$ and $(B)$ are correct.
As an alternative,
$$ I \geq \int_{0}^{\pi/2}\frac{dx}{1-\frac{x^2}{\pi^2}}=\pi\cdot\text{arctanh}\left(\frac{1}{2}\right)=\frac{\pi}{2}\log(3)>\frac{\pi}{2}\tag{4}$$
and:
$$ I \leq \int_{0}^{\pi/2}\frac{x}{\frac{2}{\pi}\,x}\,dx = \frac{\pi^2}{4}.\tag{5}$$
A: From the hint I gave in the comment you can say that the function is $f(x)=\frac{x}{\sin x}$. The maximum and minimum values that $f(x)$ attains in the given interval are $\pi/2$ and $1$ respectively. Now can you apply this to get upper and lower bounds for the integral?
EDIT: As Hagen has pointed out, the differential equation gives us the general form $f(x)=\dfrac{x+c}{\sin x}$. But the boundedness condition mandates that $c=0$, so that's that.
