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Question Number 65691 by mathmax by abdo last updated on 02/Aug/19

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{{cos}^{2}x}{cosx+sinx}dx$$

Commented by mathmax by abdo last updated on 02/Aug/19

$$tan\left(\frac{u}{2}\right)methodletI={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{2}x}{cosx+sinx}dx\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{cosx}{1+tanx}dx$$$$changementtan\left(\frac{x}{2}\right)=tgiveI={\int }_0^1\frac{\frac{1-t^2}{1+t^2}}{1+\frac{2t}{1-t^2}}\frac{2dt}{1+t^2}$$$$=2{\int }_0^1\frac{1-t^2}{{(1+t^2)}^2}\frac{1-t^2}{1-t^2+2t}dt=2{\int }_0^1\frac{{(1-t^2)}^2}{(1+t^2)(-t^2+2t+1)}dt$$$$=-2{\int }_0^1\frac{{(1-t^2)}^2}{(1+t^2)(t^2-2t-1)}dtletdecompose$$$$F\left(t\right)=\frac{t^4-2t^2+1}{(t^2+1)(t^2-2t-1)}\Rightarrow F\left(t\right)=\frac{t^4-2t^2+1}{t^4-2t^3-t^2+t^2-2t-1}$$$$=\frac{t^4-2t^2+1}{t^4-2t^3-2t-1}\Rightarrow F\left(t\right)-1=\frac{t^4-2t^2+1-t^4+2t^3+2t+1}{t^4-2t^3-2t-1}$$$$=\frac{2t^3-2t^2+2t+2}{(t^2+1)(t^2-2t-1)}=G\left(t\right)=\frac{at+b}{t^2+1}+\frac{c}{t-t_1}+\frac{d}{t-t_2}$$$${\mathrm{\Delta }}^{{}^{\prime }}=1+1=2\Rightarrow t_1=1+\sqrt{2}and{t}_2=1-\sqrt{2}$$$$c={lim}_{t\to t_1}(t-t_1)G\left(t\right)=\frac{2t_1^3-2t_1^2+2t_1+2}{(t_1^2+1)(t_1-t_2)}(t_1-t_2=2\sqrt{2})$$$$d={lim}_{t\to t_2}(t-t_2)G\left(t_2\right)=\frac{2t_2^3-2t_2^2+2t_2+2}{(t_2^2+1)(t_2-t_1)}$$$${lim}_{t\to +\mathrm{\infty }}tG\left(t\right)=2=a+c+d\Rightarrow a=2-c-d$$$$G\left(0\right)=-2=b-\frac{c}{t_1}-\frac{d}{t_2}\Rightarrow b=\frac{c}{t_1}+\frac{d}{t_2}-2resttosimplifythe$$$$calculus\Rightarrow \int F\left(t\right)dt=t+\int G\left(t\right)dt$$$$=t+cln\mid t-t_1\mid +d\mid t-t_2\mid +\frac{a}{2}ln(t^2+1)+barctant+\lambda$$$$\dots .becontinued\dots .$$

Answered by Tanmay chaudhury last updated on 02/Aug/19

$$I={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{2}x}{cosx+sinx}dx$$$$I={\int }_0^{\frac{\pi }{2}}\frac{{cos}^2(\frac{\pi }{2}-x)}{cos(\frac{\pi }{2}-x)+sin(\frac{\pi }{2}-x)}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{2}x}{sinx+cosx}dx$$$$2I={\int }_0^{\frac{\pi }{2}}\frac{dx}{sinx+cosx}$$$$2I={\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}(\frac{1}{\sqrt{2}}sinx+\frac{1}{\sqrt{2}}cosx)}$$$$I=\frac{1}{2\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{dx}{sin(\frac{\pi }{4}+x)}$$$$I=\frac{1}{2\sqrt{2}}{\int }_0^{\frac{\pi }{4}}cosec(\frac{\pi }{4}+x)dx$$$$I=\frac{1}{2\sqrt{2}}\mid lntan\left(\frac{\frac{\pi }{4}+x}{2}\right){\mid }_0^{\frac{\pi }{4}}$$$$=\frac{1}{2\sqrt{2}}[lntan\left(\frac{\pi }{4}\right)-lntan\left(\frac{\pi }{8}\right)]$$$$=\frac{-1}{2\sqrt{2}}lntan\left(\frac{\pi }{8}\right)$$

Commented by mathmax by abdo last updated on 02/Aug/19

$$thankyousirtanmay$$

Commented by Tanmay chaudhury last updated on 02/Aug/19

$$mosteelcomesir\dots$$

Commented by peter frank last updated on 02/Aug/19

$$thankyou$$

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