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Question Number 29443 by prof Abdo imad last updated on 08/Feb/18

find ∫_0 ^π     ((sinx)/(√(1+sin^2 x)))dx

$${find}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{sinx}}{\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx} \\ $$

Answered by sma3l2996 last updated on 09/Feb/18

t=cosx⇒dt=−sinxdx  ∫_0 ^π ((sinx)/(√(1+sin^2 x)))dx=∫_(−1) ^1 (dt/(√(1+(1−t^2 ))))=∫_(−1) ^1 (dt/((√2)(√(1−((t/(√2)))^2 ))))  t=(√2)u⇒dt=(√2)du  ∫_0 ^π ((sinx)/(√(1+sin^2 x)))dx=∫_(−(√2)/2) ^((√2)/2) (du/(√(1−u^2 )))  u=siny⇒dy=(du/(√(1−u^2 )))  ∫_0 ^π ((sinx)/(√(1+sin^2 x)))dx=∫_(−π/4) ^(π/4) dy=(π/2)

$${t}={cosx}\Rightarrow{dt}=−{sinxdx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\sqrt{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }} \\ $$$${t}=\sqrt{\mathrm{2}}{u}\Rightarrow{dt}=\sqrt{\mathrm{2}}{du} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx}=\int_{−\sqrt{\mathrm{2}}/\mathrm{2}} ^{\sqrt{\mathrm{2}}/\mathrm{2}} \frac{{du}}{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }} \\ $$$${u}={siny}\Rightarrow{dy}=\frac{{du}}{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {x}}}{dx}=\int_{−\pi/\mathrm{4}} ^{\pi/\mathrm{4}} {dy}=\frac{\pi}{\mathrm{2}} \\ $$

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