Integration Questions

Question Number 17011 by arnabpapu550@gmail.com last updated on 29/Jun/17

$$\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{x}\sqrt{\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\$$

Answered by sma3l2996 last updated on 29/Jun/17

$${I}=\int_{\mathrm{0}} ^{{a}} \frac{{x}.\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{dx} \\$$$${t}^{\mathrm{2}} ={a}^{\mathrm{2}} −{x}^{\mathrm{2}} \Rightarrow{tdt}=−{xdx} \\$$$${I}=−\int_{{a}} ^{\mathrm{0}} \frac{{t}}{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{dt}=\left[\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }\right]_{{a}} ^{\mathrm{0}} \\$$$${I}={a}\sqrt{\mathrm{2}}−{a}\Leftrightarrow{I}={a}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\$$

Commented by arnabpapu550@gmail.com last updated on 30/Jun/17

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\Pi}{\mathrm{2}}−\mathrm{1}\right) \\$$

Commented by sma3l2996 last updated on 30/Jun/17

$${I}=\int_{{a}} ^{\mathrm{0}} \frac{{t}}{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\left(−{tdt}\right)=\int_{\mathrm{0}} ^{{a}} \frac{{t}^{\mathrm{2}} }{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{dt} \\$$$$=−\int_{\mathrm{0}} ^{{a}} \frac{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} }{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{dt}=\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{{a}} \frac{{dt}}{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }{dt} \\$$$$=\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{{a}} \frac{{dt}}{\sqrt{\mathrm{2}}{a}.\sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}}{a}}\right)^{\mathrm{2}} }}−\sqrt{\mathrm{2}}{a}\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}}{a}}\right)^{\mathrm{2}} }{dt} \\$$$${y}=\frac{{t}}{\sqrt{\mathrm{2}}{a}}\Rightarrow{dt}=\sqrt{\mathrm{2}}{a}.{dy} \\$$$${I}=\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \frac{{dy}}{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}−\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \sqrt{\mathrm{1}−{y}^{\mathrm{2}} }{dy} \\$$$${I}=\mathrm{2}{a}^{\mathrm{2}} \left[{sin}^{−\mathrm{1}} {y}\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} −\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \sqrt{\mathrm{1}−{y}^{\mathrm{2}} }{dy} \\$$$${sin}\left({u}\right)=\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\Rightarrow{cos}\left({u}\right){du}=\frac{−{ydy}}{{sin}\left({u}\right)} \\$$$${sin}\left({u}\right).\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({u}\right)}{du}=−{ydy} \\$$$${sin}\left({u}\right).\sqrt{\mathrm{1}−\mathrm{1}+{y}^{\mathrm{2}} }{du}=−{ydy}\Leftrightarrow{sin}\left({u}\right).{y}.{du}=−{ydy} \\$$$${sin}\left({u}\right){du}=−{dy} \\$$$${I}=\mathrm{2}{a}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}−\mathrm{2}{a}^{\mathrm{2}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} {sin}\left({u}\right).\left(−{sin}\left({u}\right){du}\right) \\$$$$={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+\mathrm{2}{a}^{\mathrm{2}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} {sin}^{\mathrm{2}} \left({u}\right){du}={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+{a}^{\mathrm{2}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{u}\right)\right){du} \\$$$$={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+{a}^{\mathrm{2}} \left[{u}−\frac{{sin}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} ={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+{a}^{\mathrm{2}} \left(−\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)={a}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$${I}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right) \\$$