Question Number 51989 by maxmathsup by imad last updated on 01/Jan/19
$${calculate}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{sinx}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by peter frank last updated on 01/Jan/19
![∫((sin x)/(2−cos^2 x))dx u=cos x du=−sin xdx dx=−(du/(sin x)) ∫((sin x)/(2−u^2 )).−(du/(sin x)) ∫−(du/(2−u^2 )) −(1/(2(√2)))[∫(du/( (√2)+u))−∫(du/( (√2) −u))] −(1/(2(√(2 ))))ln[(((√2)+u)/( (√2) −u))]+B −(1/(2(√(2 ))))ln[(((√2)+cos x)/( (√2) −cos x))]+B](https://www.tinkutara.com/question/Q52000.png)
$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} {x}}{dx} \\ $$$${u}=\mathrm{cos}\:{x} \\ $$$${du}=−\mathrm{sin}\:{xdx} \\ $$$${dx}=−\frac{{du}}{\mathrm{sin}\:{x}} \\ $$$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{2}−{u}^{\mathrm{2}} }.−\frac{{du}}{\mathrm{sin}\:{x}} \\ $$$$\int−\frac{{du}}{\mathrm{2}−{u}^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\int\frac{{du}}{\:\sqrt{\mathrm{2}}+{u}}−\int\frac{{du}}{\:\sqrt{\mathrm{2}}\:−{u}}\right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}}{ln}\left[\frac{\sqrt{\mathrm{2}}+{u}}{\:\sqrt{\mathrm{2}}\:−{u}}\right]+{B} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}}{ln}\left[\frac{\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2}}\:−\mathrm{cos}\:{x}}\right]+{B} \\ $$$$ \\ $$