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=tan θ x>1 ⇒ nπ+(π/4) < θ < nπ+(π/2) ⇒ ((2x)/(1+x^2 )) = sin 2θ 2nπ+(π/2) < 2θ < 2nπ+π sin^(−1) (sin 2θ) = sin^(−1) sin (π−2θ) 2mπ < π−2θ < 2mπ+(π/2) m=0 is appropriate since π−2θ should lie in [−(π/2), (π/2)] sin^(−1) sin (π−2θ)= π−2θ = π−2tan^(−1) x I=∫(π−2tan^(−1) x)dx =πx−2(xtan^(−1) x−(1/2)∫ ((2xdx)/(1+x^2 )) )+c I=πx−2xtan^(−1) x+ln ∣1+x^2 ∣+c .](Q37543.png)
$${let}\:{x}=\mathrm{tan}\:\theta \\ $$ $$\:{x}>\mathrm{1}\:\Rightarrow\:\:\:\:{n}\pi+\frac{\pi}{\mathrm{4}}\:<\:\theta\:<\:{n}\pi+\frac{\pi}{\mathrm{2}} \\ $$ $$\Rightarrow\:\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\mathrm{sin}\:\mathrm{2}\theta \\ $$ $$\:\:\:\:\:\:\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\:<\:\:\mathrm{2}\theta\:<\:\mathrm{2}{n}\pi+\pi \\ $$ $$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{2}\theta\right)\:=\:\mathrm{sin}^{−\mathrm{1}} \mathrm{sin}\:\left(\pi−\mathrm{2}\theta\right) \\ $$ $$\:\:\:\:\:\:\:\mathrm{2}{m}\pi\:<\:\pi−\mathrm{2}\theta\:<\:\mathrm{2}{m}\pi+\frac{\pi}{\mathrm{2}} \\ $$ $${m}=\mathrm{0}\:\:{is}\:{appropriate}\:{since}\: \\ $$ $$\:\:\:\:\:\pi−\mathrm{2}\theta\:{should}\:{lie}\:{in}\:\left[−\frac{\pi}{\mathrm{2}},\:\frac{\pi}{\mathrm{2}}\right] \\ $$ $$\mathrm{sin}^{−\mathrm{1}} \mathrm{sin}\:\left(\pi−\mathrm{2}\theta\right)=\:\pi−\mathrm{2}\theta \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\pi−\mathrm{2tan}^{−\mathrm{1}} {x} \\ $$ $${I}=\int\left(\pi−\mathrm{2tan}^{−\mathrm{1}} {x}\right){dx} \\ $$ $$\:\:=\pi{x}−\mathrm{2}\left({x}\mathrm{tan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{xdx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)+{c} \\ $$ $$\:{I}=\pi{x}−\mathrm{2}{x}\mathrm{tan}^{−\mathrm{1}} {x}+\mathrm{ln}\:\mid\mathrm{1}+{x}^{\mathrm{2}} \mid+{c}\:. \\ $$
Commented byrahul 19 last updated on 14/Jun/18
thank you sir. ����
Commented byrahul 19 last updated on 14/Jun/18
$$\mathrm{Ajfour}\:\mathrm{sir}\:\mathrm{pls}\:\mathrm{try}\:\mathrm{these}\:\mathrm{ques}... \\ $$ $$\mathrm{Q}.\:\mathrm{no}.\:\mathrm{36700}\:\&\:\mathrm{36096}. \\ $$