Question Number 48955 by Abdulhafeez Abu qatada last updated on 30/Nov/18
$${Evaluate}\:\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}} {\int}}\:\frac{\mathrm{2}{x}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}\:{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18
![2x=sinθ 2dx=cosθdθ ∫_0 ^(π/3) ((sinθ×θ)/( (√(1−sin^2 θ))))×((cosθdθ)/2) (1/2)∫_0 ^(π/3) θsinθdθ I=∫θsinθdθ =θ∫sinθdθ−∫[(dθ/dθ)∫sinθdθ] dθ =−θcosθ−∫−cosθdθ =−θcosθ+sinθ+c so required intregal is (1/2)∣−θcosθ+sinθ∣_0 ^(π/3) =(1/2)[(−(π/3)×(1/2)+((√3)/2))] =((−π)/(12))+((√3)/4)](https://www.tinkutara.com/question/Q48961.png)
$$\mathrm{2}{x}={sin}\theta\:\:\:\mathrm{2}{dx}={cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{{sin}\theta×\theta}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta}}×\frac{{cos}\theta{d}\theta}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\theta{sin}\theta{d}\theta \\ $$$${I}=\int\theta{sin}\theta{d}\theta \\ $$$$=\theta\int{sin}\theta{d}\theta−\int\left[\frac{{d}\theta}{{d}\theta}\int{sin}\theta{d}\theta\right]\:{d}\theta \\ $$$$=−\theta{cos}\theta−\int−{cos}\theta{d}\theta \\ $$$$=−\theta{cos}\theta+{sin}\theta+{c} \\ $$$${so}\:{required}\:{intregal}\:{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid−\theta{cos}\theta+{sin}\theta\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(−\frac{\pi}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right] \\ $$$$=\frac{−\pi}{\mathrm{12}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$
Answered by MJS last updated on 30/Nov/18
$${u}'=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}\:\Rightarrow\:{u}=−\frac{\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$${v}=\mathrm{arcsin}\:\mathrm{2}{x}\:\Rightarrow\:{v}'=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int{u}'{v}={uv}−\int{uv}' \\ $$$$\int\frac{\mathrm{2}{x}\mathrm{arcsin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}={x}−\frac{\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}}\mathrm{arcsin}\:\mathrm{2}{x} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}} {\int}}\:\frac{\mathrm{2}{x}\mathrm{arcsin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}\:{dx}=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\pi}{\mathrm{12}} \\ $$
Answered by Abdulhafeez Abu qatada last updated on 02/Dec/18
