Question Number 143071 by cesarL last updated on 09/Jun/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{8}{dx}}{{tgx}+\mathrm{1}} \\ $$
Answered by TheSupreme last updated on 09/Jun/21
$${y}={tan}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}={dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{8}{dy}}{\left({y}+\mathrm{1}\right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{8}}{\left({y}+\mathrm{1}\right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}=\frac{{A}}{{y}+\mathrm{1}}+\frac{{By}+{C}}{{y}^{\mathrm{2}} +\mathrm{1}}=\frac{{Ay}^{\mathrm{2}} +{A}+{By}^{\mathrm{2}} +{Cy}+{By}+{C}}{{D}}=\frac{\mathrm{8}}{{D}} \\ $$$${A}+{B}=\mathrm{0} \\ $$$${C}+{B}=\mathrm{0} \\ $$$${A}+{C}=\mathrm{8} \\ $$$${A}={C}=−{B}=\mathrm{4} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}}{{y}+\mathrm{1}}+\frac{−\mathrm{4}{y}+\mathrm{4}}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}=\mathrm{4}{ln}\left({y}+\mathrm{1}\right)−\mathrm{2}{ln}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)+\mathrm{4}{arctan}\left({y}\right)= \\ $$$$=\mathrm{4}{ln}\left(\mathrm{2}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)+\mathrm{4}\frac{\pi}{\mathrm{4}}=\mathrm{2}{ln}\left(\mathrm{2}\right)−\pi \\ $$
Answered by Olaf_Thorendsen last updated on 09/Jun/21
![C = ∫_0 ^(π/4) ((8cosx)/(cosx+sinx)) dx = ∫_0 ^(π/4) (8/(tanx+1)) dx S = ∫_0 ^(π/4) ((8sinx)/(cosx+sinx)) dx C+S = ∫_0 ^(π/4) 8dx = 2π (1) C−S = 8∫_0 ^(π/4) ((cosx−sinx)/(cosx+sinx)) dx C−S = 8[ln∣cox+sinx∣]_0 ^(π/4) C−S = 8ln(√2) = 4ln2 (2) (((1)+(2))/2) : C = π+2ln2](https://www.tinkutara.com/question/Q143079.png)
$$\mathrm{C}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{8cos}{x}}{\mathrm{cos}{x}+\mathrm{sin}{x}}\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{8}}{\mathrm{tan}{x}+\mathrm{1}}\:{dx} \\ $$$$\mathrm{S}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{8sin}{x}}{\mathrm{cos}{x}+\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{C}+\mathrm{S}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{8}{dx}\:=\:\mathrm{2}\pi\:\left(\mathrm{1}\right) \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{cos}{x}−\mathrm{sin}{x}}{\mathrm{cos}{x}+\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\mathrm{8}\left[\mathrm{ln}\mid\mathrm{co}{x}+\mathrm{sin}{x}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\mathrm{8ln}\sqrt{\mathrm{2}}\:=\:\mathrm{4ln2}\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)+\left(\mathrm{2}\right)}{\mathrm{2}}\::\:\mathrm{C}\:=\:\pi+\mathrm{2ln2} \\ $$