Question Number 145803 by Mrsof last updated on 08/Jul/21
Commented by Mrsof last updated on 08/Jul/21
$${help}\:{me}\:{sir}\:{by}\:{reman}\:{intigrable} \\ $$
Answered by mathmax by abdo last updated on 08/Jul/21
![∣cosx∣≤1 ⇒∣∫_a ^b ((cosx)/x^2 )dx∣≤∫_a ^b ((∣cosx∣)/x^2 )dx≤∫_a ^b (dx/x) and ∫_a ^b (dx/x^2 )=[−(1/x)]_a ^b =(1/a)−(1/b) (if a<b)](Q145842.png)
$$\mid\mathrm{cosx}\mid\leqslant\mathrm{1}\:\Rightarrow\mid\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\mid\leqslant\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mid\mathrm{cosx}\mid}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\leqslant\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\mathrm{and}\:\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} }=\left[−\frac{\mathrm{1}}{\mathrm{x}}\right]_{\mathrm{a}} ^{\mathrm{b}} \:=\frac{\mathrm{1}}{\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{b}}\:\:\:\left(\mathrm{if}\:\mathrm{a}<\mathrm{b}\right) \\ $$