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Question Number 143071 by cesarL last updated on 09/Jun/21

$${\int }_0^{\frac{\pi }{4}}\frac{8dx}{tgx+1}$$

Answered by TheSupreme last updated on 09/Jun/21

$$y=tan\left(x\right)$$$$\frac{1}{1+y^2}dy=dx$$$${\int }_0^1\frac{8dy}{(y+1)(1+y^2)}$$$$\frac{8}{(y+1)(1+y^2)}=\frac{A}{y+1}+\frac{By+C}{y^2+1}=\frac{{Ay}^2+A+{By}^2+Cy+By+C}{D}=\frac{8}{D}$$$$A+B=0$$$$C+B=0$$$$A+C=8$$$$A=C=-B=4$$$${\int }_0^1\frac{4}{y+1}+\frac{-4y+4}{1+y^2}dy=4ln(y+1)-2ln(1+y^2)+4arctan\left(y\right)=$$$$=4ln\left(2\right)-2ln\left(2\right)+4\frac{\pi }{4}=2ln\left(2\right)-\pi$$

Answered by Olaf_Thorendsen last updated on 09/Jun/21

$$\mathrm{C}={\int }_0^{\frac{\pi }{4}}\frac{8\cos x}{\cos x+\sin x}dx={\int }_0^{\frac{\pi }{4}}\frac{8}{\tan x+1}dx$$$$\mathrm{S}={\int }_0^{\frac{\pi }{4}}\frac{8\sin x}{\cos x+\sin x}dx$$$$\mathrm{C}+\mathrm{S}={\int }_0^{\frac{\pi }{4}}8dx=2\pi \left(1\right)$$$$\mathrm{C}-\mathrm{S}=8{\int }_0^{\frac{\pi }{4}}\frac{\cos x-\sin x}{\cos x+\sin x}dx$$$$\mathrm{C}-\mathrm{S}=8{[\ln \mid \mathrm{co}x+\sin x\mid ]}_0^{\frac{\pi }{4}}$$$$\mathrm{C}-\mathrm{S}=8\ln \sqrt{2}=4\ln 2\left(2\right)$$$$\frac{\left(1\right)+\left(2\right)}{2}:\mathrm{C}=\pi +2\ln 2$$

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