0-pi-cosx-2-sin-2-x-dx-

Question Number 107002 by Ar Brandon last updated on 08/Aug/20

\int_{0}^{π} \frac{cosx}{\sqrt{2 - \sin^{2} x}} dx

Answered by bemath last updated on 08/Aug/20

\underset{0}{\int^{π}} \frac{d (\sin x)}{\sqrt{2 - \sin^{2} x}} = \underset{\sqrt{2}}{\int^{\sqrt{2}}} \frac{d u}{\sqrt{2 - u^{2}}} = 0

Answered by mathmax by abdo last updated on 08/Aug/20

I = \int_{0}^{π} \frac{cosx}{\sqrt{2 - \sin^{2} x}} dx we have by changement sinx = t

\int \frac{cosx dx}{\sqrt{2 - \sin^{2} x}} dx = \int \frac{dt}{\sqrt{2 - t^{2}}} = \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{1 - {(\frac{t}{\sqrt{2}})}^{2}}}

=_{\frac{t}{\sqrt{2}} = u} \frac{1}{\sqrt{2}} \int \frac{\sqrt{2} du}{\sqrt{1 - u^{2}}} = \arcsin (\frac{t}{\sqrt{2}}) + c \Rightarrow I = {[\arcsin (\frac{sinx}{\sqrt{2}})]}_{0}^{π}

= 0

Answered by Ar Brandon last updated on 08/Aug/20

I = \int_{0}^{\frac{π}{2}} \frac{cosx}{\sqrt{2 - \sin^{2} x}} dx - \int_{0}^{\frac{π}{2}} \frac{cosx}{\sqrt{2 - \sin^{2} x}} dx = 0