Question Number 84556 by jagoll last updated on 14/Mar/20
$$\int\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2x}+\mathrm{2}}{\:\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{13}}}\right)\:\mathrm{dx} \\ $$
Commented by john santu last updated on 14/Mar/20
![∫ sin^(−1) (((2x+2)/( (√((2x+2)^2 +9))))) dx let 2x+2 = 3 tan θ ⇒2dx = 3sec^2 θ dθ ∫ sin^(−1) (((3tan θ)/(3sec θ))) ×(3/2) sec^2 θ dθ = ∫(3/2) sin^(−1) (sin θ) d(tan θ) = (3/2)∫ θ d(tan θ) = (3/2)[ θ tan θ −∫ tan θ dθ ] = (3/2) [ θ tan θ +∫ ((d(cos θ))/(cos θ)) ] = (3/2)[ θ tan θ + ln ∣ cos θ ∣ ]+ c = (3/2)[(((2x+2)/3))tan^(−1) (((2x+2)/2)) + ln∣(3/( (√(4x^2 +8x+13))))∣ ] + c](https://www.tinkutara.com/question/Q84587.png)
$$\int\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2x}+\mathrm{2}}{\:\sqrt{\left(\mathrm{2x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{9}}}\right)\:\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{2x}+\mathrm{2}\:=\:\mathrm{3}\:\mathrm{tan}\:\theta\:\Rightarrow\mathrm{2dx}\:=\:\mathrm{3sec}\:^{\mathrm{2}} \theta\:\mathrm{d}\theta \\ $$$$\int\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3tan}\:\theta}{\mathrm{3sec}\:\theta}\right)\:×\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{sec}\:^{\mathrm{2}} \theta\:\mathrm{d}\theta\:= \\ $$$$\int\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\theta\right)\:\mathrm{d}\left(\mathrm{tan}\:\theta\right)\:= \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\int\:\theta\:\mathrm{d}\left(\mathrm{tan}\:\theta\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\left[\:\theta\:\mathrm{tan}\:\theta\:−\int\:\mathrm{tan}\:\theta\:\mathrm{d}\theta\:\right] \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\:\left[\:\theta\:\mathrm{tan}\:\theta\:+\int\:\frac{\mathrm{d}\left(\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta}\:\right]\: \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\left[\:\theta\:\mathrm{tan}\:\theta\:+\:\mathrm{ln}\:\mid\:\mathrm{cos}\:\theta\:\mid\:\right]+\:\mathrm{c}\: \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\left[\left(\frac{\mathrm{2x}+\mathrm{2}}{\mathrm{3}}\right)\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2x}+\mathrm{2}}{\mathrm{2}}\right)\:+\:\mathrm{ln}\mid\frac{\mathrm{3}}{\:\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{13}}}\mid\:\right]\:+\:\mathrm{c} \\ $$
Answered by mind is power last updated on 14/Mar/20
$$=\int{sin}^{−} \left(\frac{\mathrm{2}{x}+\mathrm{2}}{\:\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }}\right)=\int{sin}^{−} \left(\frac{\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }}\right) \\ $$$${sin}^{−} \left(\frac{{r}}{\:\sqrt{{r}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$$${r}={tg}\left(\theta\right)\Rightarrow{sin}^{−} \left(\frac{{tg}\left(\theta\right)}{\:\sqrt{\mathrm{1}+{tg}^{\mathrm{2}} \left(\theta\right)}}\right)={sin}^{−} \left({sin}\left(\theta\right)\right)=\theta={arctan}\left({r}\right) \\ $$$$\Rightarrow\int{sin}^{−} \left(\frac{\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }}\right){dx}=\int{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}}\right){dx}\:{by}\:{part}\:{easy}\:{now} \\ $$
Commented by jagoll last updated on 14/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{can}\:\mathrm{try} \\ $$