Question Number 158973 by mnjuly1970 last updated on 11/Nov/21
Answered by qaz last updated on 11/Nov/21
![∫_0 ^(π/2) (x/((1+sin x)^2 ))dx =(π/2)∫_0 ^(π/2) (dx/((1+cos x)^2 ))−∫_0 ^(π/2) (x/((1+cos x)^2 ))dx =(π/4)∫_0 ^(π/4) (dx/(cos^4 x))−∫_0 ^(π/4) (x/(cos^4 x))dx =(π/4)∫_0 ^(π/4) (1+tan^2 x)d(tan x)−∫_0 ^(π/4) xd(tan x+(1/3)tan^3 x) =(π/4)∙[tan x+(1/3)tan^3 x]_0 ^(π/4) −x[tan x+(1/3)tan^3 x]_0 ^(π/4) +∫_0 ^(π/4) (tan x+(1/3)tan^3 x)dx =(π/3)−(π/3)+∫_0 ^(π/4) tan x((2/3)+(1/3)sec^2 x)dx =(1/3)ln2+(1/6) ⇒K=(((1/3)ln2+(1/6))/((π/3)−(1/3)ln2−(1/6)))=((2ln2+1)/(2π−2ln2−1))](https://www.tinkutara.com/question/Q159014.png)
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{\mathrm{dx}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}−\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{\mathrm{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)−\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{xd}\left(\mathrm{tan}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}\centerdot\left[\mathrm{tan}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} −\mathrm{x}\left[\mathrm{tan}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} +\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \left(\mathrm{tan}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\right)\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{3}}+\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}\:\mathrm{x}\left(\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{K}=\frac{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{6}}}{\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{6}}}=\frac{\mathrm{2ln2}+\mathrm{1}}{\mathrm{2}\pi−\mathrm{2ln2}−\mathrm{1}} \\ $$