Question Number 29443 by prof Abdo imad last updated on 08/Feb/18
$${find}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{sinx}}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx} \\ $$
Answered by sma3l2996 last updated on 09/Feb/18
$${t}={cosx}\Rightarrow{dt}=−{sinxdx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−\left(\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }} \\ $$$${t}=\sqrt{\mathrm{2}}{u}\Rightarrow{dt}=\sqrt{\mathrm{2}}{du} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx}=\int_{−\sqrt{\mathrm{2}}/\mathrm{2}} ^{\sqrt{\mathrm{2}}/\mathrm{2}} \frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }} \\ $$$${u}={siny}\Rightarrow{dy}=\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx}=\int_{−\pi/\mathrm{4}} ^{\pi/\mathrm{4}} {dy}=\frac{\pi}{\mathrm{2}} \\ $$