Question Number 37540 by rahul 19 last updated on 14/Jun/18
$$\mathrm{For}\:{x}>\mathrm{1}\:,\: \\ $$$$\int\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\:=\:? \\ $$
Answered by ajfour last updated on 14/Jun/18
$${let}\:{x}=\mathrm{tan}\:\theta \\ $$$$\:{x}>\mathrm{1}\:\Rightarrow\:\:\:\:{n}\pi+\frac{\pi}{\mathrm{4}}\:<\:\theta\:<\:{n}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\:<\:\:\mathrm{2}\theta\:<\:\mathrm{2}{n}\pi+\pi \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{2}\theta\right)\:=\:\mathrm{sin}^{−\mathrm{1}} \mathrm{sin}\:\left(\pi−\mathrm{2}\theta\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{m}\pi\:<\:\pi−\mathrm{2}\theta\:<\:\mathrm{2}{m}\pi+\frac{\pi}{\mathrm{2}} \\ $$$${m}=\mathrm{0}\:\:{is}\:{appropriate}\:{since}\: \\ $$$$\:\:\:\:\:\pi−\mathrm{2}\theta\:{should}\:{lie}\:{in}\:\left[−\frac{\pi}{\mathrm{2}},\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{sin}^{−\mathrm{1}} \mathrm{sin}\:\left(\pi−\mathrm{2}\theta\right)=\:\pi−\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\pi−\mathrm{2tan}^{−\mathrm{1}} {x} \\ $$$${I}=\int\left(\pi−\mathrm{2tan}^{−\mathrm{1}} {x}\right){dx} \\ $$$$\:\:=\pi{x}−\mathrm{2}\left({x}\mathrm{tan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{xdx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)+{c} \\ $$$$\:{I}=\pi{x}−\mathrm{2}{x}\mathrm{tan}^{−\mathrm{1}} {x}+\mathrm{ln}\:\mid\mathrm{1}+{x}^{\mathrm{2}} \mid+{c}\:. \\ $$
Commented by rahul 19 last updated on 14/Jun/18
thank you sir.
Commented by rahul 19 last updated on 14/Jun/18
$$\mathrm{Ajfour}\:\mathrm{sir}\:\mathrm{pls}\:\mathrm{try}\:\mathrm{these}\:\mathrm{ques}… \\ $$$$\mathrm{Q}.\:\mathrm{no}.\:\mathrm{36700}\:\&\:\mathrm{36096}. \\ $$