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Question Number 35054 by math khazana by abdo last updated on 14/May/18

$$find{\int }_0^{\frac{\pi }{4}}\frac{xdx}{2+cosx}$$

Commented by prof Abdo imad last updated on 16/May/18

$$changementtan\left(\frac{x}{2}\right)=tgive$$$$I={\int }_0^{\sqrt{2}-1}\frac{2arctant}{2+\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$$$=4{\int }_0^{\sqrt{2}-1}\frac{arctant}{2+2t^2+1-t^2}dt$$$$=4{\int }_0^{\sqrt{2}-1}\frac{arctan\left(t\right)}{3+t^2}dt$$$${=}_{t=\sqrt{3}u}4{\int }_0^{\frac{\sqrt{2}-1}{3}}\frac{arctan\left(\sqrt{3}u\right)}{3(1+u^2)}\sqrt{3}du$$$$=\frac{4\sqrt{3}}{3}{\int }_0^{\frac{\sqrt{2}-1}{3}}\frac{arctan\left(\sqrt{3}u\right)}{1+u^2}dubyparts$$$${\int }_0^{\frac{\sqrt{2}-1}{3}}\frac{arctan\left(\sqrt{3}u\right)}{1+u^2}du$$$$={\left[arctan\left(u\right)arctan\left(\sqrt{3}u\right)\right]}_0^{\frac{\sqrt{2}-1}{3}}-\sqrt{3}{\int }_0^{\frac{\sqrt{2}-1}{3}}\frac{arctan\left(u\right)}{1+3u^2}du$$$$=arctan\left(\frac{\sqrt{2}-1}{3}\right)arctan\left(\sqrt{3}\frac{\sqrt{2}-1}{3}\right)$$$$-\sqrt{3}{\int }_0^{\frac{\sqrt{2}-1}{3}}\frac{arctan\left(u\right)}{1+3u^2}du....becontinued...$$$$$$

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