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Question Number 42394 by abdo.msup.com last updated on 24/Aug/18

$$find{\int }_0^1\frac{dt}{t+\sqrt{1-t^2}}dt$$

Commented by maxmathsup by imad last updated on 25/Aug/18

$$changementt=sinxgive$$$$I={\int }_0^{\frac{\pi }{2}}\frac{cosxdx}{sinx+cosx}dx={\int }_0^{\frac{\pi }{2}}\frac{dx}{tanx+1}$$$$I{=}_{tanx=u}{\int }_0^{+\mathrm{\infty }}\frac{1}{1+u}\frac{du}{1+u^2}={\int }_0^{\mathrm{\infty }}\frac{du}{(u+1)(u^2+1)}letdecompose$$$$F\left(u\right)=\frac{1}{(u+1)(u^2+1)}\Rightarrow F\left(u\right)=\frac{a}{u+1}+\frac{bu+c}{u^2+1}$$$$a={lim}_{u\to -1}(u+1)F\left(u\right)=\frac{1}{2}$$$${lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=0=a+b\Rightarrow b=-\frac{1}{2}\Rightarrow F\left(u\right)=\frac{1}{2(u+1)}+\frac{-\frac{1}{2}u+c}{u^2+1}$$$$F\left(0\right)=1=\frac{1}{2}+c\Rightarrow c=\frac{1}{2}\Rightarrow F\left(u\right)=\frac{1}{2(u+1)}-\frac{1}{2}\frac{u-1}{u^2+1}\Rightarrow$$$$I={\int }_0^{\mathrm{\infty }}F\left(u\right)du=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{du}{u+1}-\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{2u-2}{u^2+1}du$$$$={[\frac{1}{2}ln\mid u+1\mid -\frac{1}{4}ln(u^2+1)]}_0^{+\mathrm{\infty }}+\frac{1}{2}{\left[arctanu\right]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{2}{[ln\mid \frac{u+1}{\sqrt{u^2+1}}\mid ]}_0^{+\mathrm{\infty }}+\frac{\pi }{4}=0+\frac{\pi }{4}=\frac{\pi }{4}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

$$t=sin\alpha dt=cos\alpha d\alpha$$$${\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{cos\alpha }{cos\alpha +sin\alpha }d\alpha$$$$=\frac{1}{2}{\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{cos\alpha +sin\alpha +cos\alpha -sin\alpha }{cos\alpha +sin\alpha }d\alpha$$$$=\frac{1}{2}{\int }_0^{\frac{\mathrm{\Pi }}{2}}d\alpha +\frac{1}{2}{\int }_0^{\frac{\mathrm{\Pi }}{2}}\frac{d(sin\alpha +cos\alpha )}{cos\alpha +sin\alpha }$$$$=\frac{1}{2}\left(\frac{\mathrm{\Pi }}{2}\right)+\frac{1}{2}\mid ln(cos\alpha +sin\alpha ){\mid }_0^{\frac{\mathrm{\Pi }}{2}}$$$$=\frac{\mathrm{\Pi }}{4}+\frac{1}{2}\{ln\left(1\right)-ln1\}$$$$=\frac{\mathrm{\Pi }}{4}$$

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