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Question Number 51989 by maxmathsup by imad last updated on 01/Jan/19

calculate ∫_(π/4) ^(π/3)    ((sinx)/(1+sin^2 x))dx

$${calculate}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{sinx}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$

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

$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} {x}}{dx} \\ $$$${u}=\mathrm{cos}\:{x} \\ $$$${du}=−\mathrm{sin}\:{xdx} \\ $$$${dx}=−\frac{{du}}{\mathrm{sin}\:{x}} \\ $$$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{2}−{u}^{\mathrm{2}} }.−\frac{{du}}{\mathrm{sin}\:{x}} \\ $$$$\int−\frac{{du}}{\mathrm{2}−{u}^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\int\frac{{du}}{\sqrt{\mathrm{2}}+{u}}−\int\frac{{du}}{\sqrt{\mathrm{2}}\:−{u}}\right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}}{ln}\left[\frac{\sqrt{\mathrm{2}}+{u}}{\sqrt{\mathrm{2}}\:−{u}}\right]+{B} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}}{ln}\left[\frac{\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}}{\sqrt{\mathrm{2}}\:−\mathrm{cos}\:{x}}\right]+{B} \\ $$$$ \\ $$

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