Question Number 225629 by sonukgindia last updated on 05/Nov/25
Answered by Spillover last updated on 05/Nov/25
$$\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\right)? \\ $$
Answered by Spillover last updated on 05/Nov/25
![∫_b ^a f(x)dx=∫_0 ^a f(a−x)dx a=(π/2) I=∫_0 ^(π/2) log (1+(1/2)sin θ)=∫_0 ^(π/2) log (1+(1/2)cos θ) I=∫_0 ^(π/2) log (1+(1/2)sin θ) I=∫_0 ^(π/2) log (1+(1/2)cos θ) 2I=∫_0 ^(π/2) log [1+(1/2)sin θ+(1/2)cos θ+(1/4)sin θcos θ]dθ but log (1+acos θ)dx=(π/2)log (((1+(√(1−a^2 )))/2)) I=(π/2)log (((1+(√(1−((1/2))^2 )))/2)) I=(π/2)log (((2+(√3))/4))](https://www.tinkutara.com/question/Q225675.png)
$$\int_{{b}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx} \\ $$$${a}=\frac{\pi}{\mathrm{2}}\:\:\:\:\:\: \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\:\theta\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\:\theta\right) \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\:\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\theta\mathrm{cos}\:\theta\right]{d}\theta \\ $$$${but} \\ $$$$\mathrm{log}\:\left(\mathrm{1}+{a}\mathrm{cos}\:\theta\right){dx}=\frac{\pi}{\mathrm{2}}\mathrm{log}\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{2}}\right) \\ $$$${I}=\frac{\pi}{\mathrm{2}}\mathrm{log}\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\mathrm{2}}\right) \\ $$$${I}=\frac{\pi}{\mathrm{2}}\mathrm{log}\:\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$