Question Number 107002 by Ar Brandon last updated on 08/Aug/20
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cosx}}{\:\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\ $$
Answered by bemath last updated on 08/Aug/20
$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\:\sqrt{\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} {x}}}\:=\:\underset{\:\sqrt{\mathrm{2}}} {\overset{\sqrt{\mathrm{2}}} {\int}}\:\frac{{du}}{\:\sqrt{\mathrm{2}−{u}^{\mathrm{2}} }}\:=\:\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 08/Aug/20
![I =∫_0 ^π ((cosx)/( (√(2−sin^2 x))))dx we have by changement sinx =t ∫ ((cosx dx)/( (√(2−sin^2 x))))dx =∫ (dt/( (√(2−t^2 )))) =(1/( (√2)))∫ (dt/( (√(1−((t/( (√2))))^2 )))) =_((t/( (√2)))=u) (1/( (√2)))∫ (((√2)du)/( (√(1−u^2 )))) =arcsin((t/( (√2))))+c ⇒I=[arcsin(((sinx)/( (√2))))]_0 ^π =0](https://www.tinkutara.com/question/Q107038.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{cosx}}{\:\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{by}\:\mathrm{changement}\:\mathrm{sinx}\:=\mathrm{t} \\ $$$$\int\:\frac{\mathrm{cosx}\:\mathrm{dx}}{\:\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}\:=\int\:\frac{\mathrm{dt}}{\:\sqrt{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\frac{\mathrm{dt}}{\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }} \\ $$$$=_{\frac{\mathrm{t}}{\:\sqrt{\mathrm{2}}}=\mathrm{u}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\:\frac{\sqrt{\mathrm{2}}\mathrm{du}}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }}\:=\mathrm{arcsin}\left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{c}\:\Rightarrow\mathrm{I}=\left[\mathrm{arcsin}\left(\frac{\mathrm{sinx}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\mathrm{0} \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 08/Aug/20
$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cosx}}{\:\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cosx}}{\:\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx}=\mathrm{0} \\ $$