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Question Number 172681 by mnjuly1970 last updated on 30/Jun/22

Answered by Jamshidbek last updated on 30/Jun/22

$$\mathrm{Hint}:\mathrm{Lemma}:{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{dx}$$$$\mathrm{after}\mathrm{tg}\frac{\mathrm{x}}{2}=\mathrm{t}.$$

Answered by Mathspace last updated on 30/Jun/22

$$$$$$f\left(a\right)={\int }_0^{\frac{\pi }{4}}\frac{dx}{a+cosx+sinx}$$$$\Rightarrow f^{{}^{\prime }}\left(a\right)=-{\int }_0^{\frac{\pi }{4}}\frac{dx}{{(a+cosx+sinx)}^2}$$$$\Rightarrow {\int }_0^{\frac{\pi }{4}}\frac{dx}{{(a+cosx+sinx)}^2}=-f^{{}^{\prime }}\left(a\right)$$$$\Rightarrow I=-f^{{}^{\prime }}\left(\sqrt{2}\right)$$$$wetakea>1$$$$f\left(a\right){=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^{\sqrt{2}-1}\frac{2dt}{(1+t^2)(a+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2})}$$

Commented by Mathspace last updated on 30/Jun/22

$$f\left(a\right)=2{\int }_0^{\sqrt{2}-1}\frac{dt}{a+{at}^2+1-t^2+2t}$$$$=2{\int }_0^{\sqrt{2}-1}\frac{dt}{(a-1)t^2+2t+a+1}$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-(a^2-1)=2-a^2$$$$t_1=\frac{-1+\sqrt{2-a^2}}{a-1}$$$$t_2=\frac{-1-\sqrt{2-a^2}}{a-1}\Rightarrow$$$$f\left(a\right)=2{\int }_0^{\sqrt{2}-1}\frac{dt}{(a-1)(t-t_1)(t-t_2)}$$$$=\frac{2}{a-1}\times \frac{2\sqrt{2-a^2}}{a-1}{\int }_0^{\sqrt{2}-1}(\frac{1}{t-t_1}-\frac{1}{t-t_2})dt$$$$=\frac{4\sqrt{2-a^2}}{{(a-1)}^2}{[ln\mid \frac{t-t_1}{t-t_2}\mid ]}_0^{\sqrt{2}-1}$$$$=\frac{4\sqrt{2-a^2}}{{(a-1)}^2}\{ln\mid \frac{\sqrt{2}-1-t_1}{\sqrt{2}-1-t_2}\mid +ln\mid \frac{t_2}{t_1}\mid \}$$$$resttocalculate{f}^{{}^{\prime }}\left(a\right)...becontinued...$$

Answered by Mathspace last updated on 30/Jun/22

$$\mathrm{\Phi }={\int }_0^{\frac{\pi }{4}}\frac{dx}{{(\sqrt{2}+\sqrt{2}cos(x-\frac{\pi }{4}))}^2}$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{dx}{{(1+cos(x-\frac{\pi }{4}))}^2}$$$${=}_{x-\frac{\pi }{4}=-t}\frac{1}{2}{\int }_{\frac{\pi }{4}}^o\frac{-dt}{{(1+cost)}^2}$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{dt}{{(1+cost)}^2}(tan\left(\frac{t}{2}\right)=y)$$$$=\frac{1}{2}{\int }_0^{\sqrt{2}-1}\frac{2dy}{(y^2+1){(1+\frac{1-y^2}{1+y^2})}^2}$$$$={\int }_0^{\sqrt{2}-1}\frac{dy}{(y^2+1){\left(\frac{2}{1+y^2}\right)}^2}$$$$={\int }_0^{\sqrt{2}-1}\frac{1+y^2}{4}dy$$$$=\frac{1}{4}(\sqrt{2}-1)+\frac{1}{12}{\left[y^3\right]}_0^{\sqrt{2}-1}$$$$=\frac{\sqrt{2}-1}{4}+\frac{{(\sqrt{2}-1)}^3}{12}$$

Commented by Tawa11 last updated on 01/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 30/Jun/22

$$\cos x+\sin x=\sqrt{2}\cos (x-\frac{\pi }{4})$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{d(x-\frac{\pi }{4})}{2{(1+\cos (x-\frac{\pi }{4}))}^2}$$$$\mathrm{\Omega }={\int }_{-\frac{\pi }{4}}^0\frac{dt}{2{(1+\cos t)}^2}$$$$\mathrm{\Omega }=\frac{1}{4}{\int }_{-\frac{\pi }{4}}^0\frac{d\left(\frac{t}{2}\right)}{{\cos }^4\left(\frac{t}{2}\right)}$$$$\mathrm{\Omega }=\frac{1}{4}{\int }_{-\frac{\pi }{8}}^0\frac{du}{{\cos }^{4}u}$$$$\mathrm{\Omega }=\frac{1}{4}{[\tan u+\frac{{\tan }^{3}u}{3}]}_{-\frac{\pi }{8}}^0$$$$\mathrm{\Omega }=\frac{1}{4}\tan \frac{\pi }{8}(1+\frac{1}{3}\times {\tan }^2\frac{\pi }{8})$$$$\mathrm{\Omega }=\frac{\sqrt{2}-1}{4}(1+\frac{{(\sqrt{2}-1)}^2}{3})=\frac{4\sqrt{2}-5}{6}✓$$$$$$$$\tan \frac{\pi }{4}=\frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}=1$$$$\Rightarrow \tan \frac{\pi }{8}=\sqrt{2}-1$$

Commented by Tawa11 last updated on 01/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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