Evaluate-sinx-1-sin-2-x-dx-

Question Number 97782 by 175 last updated on 09/Jun/20

E v a l u a t e :

\int \frac{s i n x}{1 + {s i n}^{2} x} d x

Commented by mr W last updated on 09/Jun/20

= - \int \frac{d (\cos x)}{(\sqrt{2} - \cos x) (\sqrt{2} + \cos x)}

= - \frac{\sqrt{2}}{4} \int [\frac{1}{\sqrt{2} - \cos x} + \frac{1}{\sqrt{2} + \cos x}] d (\cos x)

= \frac{\sqrt{2}}{4} \ln \frac{\sqrt{2} - \cos x}{\sqrt{2} + \cos x} + C

Answered by niroj last updated on 09/Jun/20

\int \frac{\sin x}{1 + \sin^{2} x} dx

\int \frac{\sin x dx}{1 + 1 - \cos^{2} x} = \int \frac{\sin x dx}{2 - \cos^{2} x}

Put, \cos x = t

- \sin x dx = dt

sinxdx = - dt

- \int \frac{1}{2 - t^{2}} dt

= - \int \frac{1}{{(\sqrt{2})}^{2} - {(t)}^{2}} dt

= - \frac{1}{2 \sqrt{2}} \log \frac{\sqrt{2} + t}{\sqrt{2} - t} + C

= - \frac{1}{2 \sqrt{2}} \log \frac{\sqrt{2} + \cos x}{\sqrt{2} - \cos x} + C / / .

Answered by abdomathmax last updated on 09/Jun/20

I = \int \frac{sinx dx}{1 + \sin^{2} x} \Rightarrow I = \int \frac{sinxdx}{2 - \cos^{2} x}

we do the changement cosx = t \Rightarrow

I = \int \frac{- dt}{2 - t^{2}} = \int \frac{dt}{t^{2} - 2} = \frac{1}{2 \sqrt{2}} \int (\frac{1}{t - \sqrt{2}} - \frac{1}{t + \sqrt{2}}) dt

= \frac{1}{2 \sqrt{2}} \ln ∣ \frac{t - \sqrt{2}}{t + \sqrt{2}} ∣ + C \Rightarrow

I = \frac{1}{2 \sqrt{2}} \ln ∣ \frac{cosx - \sqrt{2}}{cosx + \sqrt{2}} ∣ + C