Question Number 182769 by cortano1 last updated on 14/Dec/22
$$\:\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{{dx}}{\mathrm{sin}\:\left({x}−\mathrm{1}\right)−\mathrm{1}}\:=?\: \\ $$
Answered by ARUNG_Brandon_MBU last updated on 14/Dec/22
![I=∫_0 ^(π/4) (dx/(sin(x−1)−1))=∫_0 ^(π/4) ((sin(x−1)+1)/(sin^2 (x−1)−1))dx =−∫_(−1) ^((π/4)−1) ((sint+1)/(cos^2 t))dt=[(1/(cost))+tant]_((π/4)−1) ^(−1) =sec(1)+tan(1)−sec((π/4)−1)−tan((π/4)−1)](https://www.tinkutara.com/question/Q182785.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{sin}\left({x}−\mathrm{1}\right)−\mathrm{1}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin}\left({x}−\mathrm{1}\right)+\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left({x}−\mathrm{1}\right)−\mathrm{1}}{dx} \\ $$$$\:\:=−\int_{−\mathrm{1}} ^{\frac{\pi}{\mathrm{4}}−\mathrm{1}} \frac{\mathrm{sin}{t}+\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} {t}}{dt}=\left[\frac{\mathrm{1}}{\mathrm{cos}{t}}+\mathrm{tan}{t}\right]_{\frac{\pi}{\mathrm{4}}−\mathrm{1}} ^{−\mathrm{1}} \\ $$$$\:\:=\mathrm{sec}\left(\mathrm{1}\right)+\mathrm{tan}\left(\mathrm{1}\right)−\mathrm{sec}\left(\frac{\pi}{\mathrm{4}}−\mathrm{1}\right)−\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\mathrm{1}\right) \\ $$