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=∫_0 ^a ((x.(√(a^2 −x^2 )))/(√(a^2 +x^2 )))dx t^2 =a^2 −x^2 ⇒tdt=−xdx I=−∫_a ^0 (t/(√(2a^2 −t^2 )))dt=[(√(2a^2 −t^2 ))]_a ^0 I=a(√2)−a⇔I=a((√2)−1)](Q17026.png)
$${I}=\int_{\mathrm{0}} ^{{a}} \frac{{x}.\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{dx} \\ $$$${t}^{\mathrm{2}} ={a}^{\mathrm{2}} −{x}^{\mathrm{2}} \Rightarrow{tdt}=−{xdx} \\ $$$${I}=−\int_{{a}} ^{\mathrm{0}} \frac{{t}}{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{dt}=\left[\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }\right]_{{a}} ^{\mathrm{0}} \\ $$$${I}={a}\sqrt{\mathrm{2}}−{a}\Leftrightarrow{I}={a}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$
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=∫_a ^0 (t/(√(2a^2 −t^2 )))(−tdt)=∫_0 ^a (t^2 /(√(2a^2 −t^2 )))dt =−∫_0 ^a ((2a^2 −t^2 −2a^2 )/(√(2a^2 −t^2 )))dt=2a^2 ∫_0 ^a (dt/(√(2a^2 −t^2 )))−∫_0 ^a (√(2a^2 −t^2 ))dt =2a^2 ∫_0 ^a (dt/((√2)a.(√(1−((t/((√2)a)))^2 ))))−(√2)a∫_0 ^a (√(1−((t/((√2)a)))^2 ))dt y=(t/((√2)a))⇒dt=(√2)a.dy I=2a^2 ∫_0 ^(1/(√2)) (dy/(√(1−y^2 )))−2a^2 ∫_0 ^(1/(√2)) (√(1−y^2 ))dy I=2a^2 [sin^(−1) y]_0 ^((√2)/2) −2a^2 ∫_0 ^((√2)/2) (√(1−y^2 ))dy sin(u)=(√(1−y^2 ))⇒cos(u)du=((−ydy)/(sin(u))) sin(u).(√(1−sin^2 (u)))du=−ydy sin(u).(√(1−1+y^2 ))du=−ydy⇔sin(u).y.du=−ydy sin(u)du=−dy I=2a^2 (π/4)−2a^2 ∫_(π/2) ^(π/4) sin(u).(−sin(u)du) =a^2 (π/2)+2a^2 ∫_(π/2) ^(π/4) sin^2 (u)du=a^2 (π/2)+a^2 ∫_(π/2) ^(π/4) (1−cos(2u))du =a^2 (π/2)+a^2 [u−((sin(2u))/2)]_(π/2) ^(π/4) =a^2 (π/2)+a^2 (−(π/4)−(1/2))=a^2 ((π/2)−(π/4)−(1/2)) I=(a^2 /2)((π/2)−1)](Q17037.png)
$${I}=\int_{{a}} ^{\mathrm{0}} \frac{{t}}{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\left(−{tdt}\right)=\int_{\mathrm{0}} ^{{a}} \frac{{t}^{\mathrm{2}} }{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{{a}} \frac{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} }{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{dt}=\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{{a}} \frac{{dt}}{\sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{{a}} \frac{{dt}}{\sqrt{\mathrm{2}}{a}.\sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}}{a}}\right)^{\mathrm{2}} }}−\sqrt{\mathrm{2}}{a}\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}}{a}}\right)^{\mathrm{2}} }{dt} \\ $$$${y}=\frac{{t}}{\sqrt{\mathrm{2}}{a}}\Rightarrow{dt}=\sqrt{\mathrm{2}}{a}.{dy} \\ $$$${I}=\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \frac{{dy}}{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}−\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \sqrt{\mathrm{1}−{y}^{\mathrm{2}} }{dy} \\ $$$${I}=\mathrm{2}{a}^{\mathrm{2}} \left[{sin}^{−\mathrm{1}} {y}\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} −\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \sqrt{\mathrm{1}−{y}^{\mathrm{2}} }{dy} \\ $$$${sin}\left({u}\right)=\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\Rightarrow{cos}\left({u}\right){du}=\frac{−{ydy}}{{sin}\left({u}\right)} \\ $$$${sin}\left({u}\right).\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({u}\right)}{du}=−{ydy} \\ $$$${sin}\left({u}\right).\sqrt{\mathrm{1}−\mathrm{1}+{y}^{\mathrm{2}} }{du}=−{ydy}\Leftrightarrow{sin}\left({u}\right).{y}.{du}=−{ydy} \\ $$$${sin}\left({u}\right){du}=−{dy} \\ $$$${I}=\mathrm{2}{a}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}−\mathrm{2}{a}^{\mathrm{2}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} {sin}\left({u}\right).\left(−{sin}\left({u}\right){du}\right) \\ $$$$={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+\mathrm{2}{a}^{\mathrm{2}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} {sin}^{\mathrm{2}} \left({u}\right){du}={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+{a}^{\mathrm{2}} \int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{u}\right)\right){du} \\ $$$$={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+{a}^{\mathrm{2}} \left[{u}−\frac{{sin}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} ={a}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}+{a}^{\mathrm{2}} \left(−\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)={a}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${I}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right) \\ $$