Question Number 132755 by liberty last updated on 16/Feb/21 | ||
Commented by KINTU last updated on 16/Feb/21 | ||
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sinx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{8}}\left(\pi−\mathrm{ln4}\right) \\ $$ | ||
Answered by EDWIN88 last updated on 16/Feb/21 | ||
$$\:\mathrm{Let}\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:,\:\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{I}+\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{and}\:\mathrm{I}−\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:=\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}} \\ $$$$\:=\:−\mathrm{ln}\:\mid\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} =\:−\mathrm{ln}\:\sqrt{\mathrm{2}}\: \\ $$$$\mathrm{We}\:\mathrm{asked}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{I}\:,\:\mathrm{we}\:\mathrm{find}\:\mathrm{it}\:\mathrm{from} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\left(\mathrm{I}+\mathrm{J}\right)+\left(\mathrm{I}−\mathrm{J}\right)\:\right] \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\pi}{\mathrm{4}}−\mathrm{ln}\:\sqrt{\mathrm{2}}\:\right]\:=\:\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$ | ||
Answered by MJS_new last updated on 16/Feb/21 | ||
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{dx}=\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\:+\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{dx}= \\ $$$$=\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\frac{−\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{dx}=\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}\right)}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{1}−\mathrm{cot}\:{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}−\mathrm{ln}\:\mid\mathrm{sin}\:{x}\mid\right]_{\pi/\mathrm{4}} ^{\pi/\mathrm{2}} = \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}} \\ $$ | ||
Answered by malwan last updated on 16/Feb/21 | ||
$$\:_{\mathrm{0}} \int^{\:\frac{\pi}{\mathrm{4}}} \frac{{sinx}\:+\left(−{cosx}+{cosx}\right)+\left({sinx}−{sinx}\right)}{{sinx}+{cosx}}{dx} \\ $$$$=\:−{ln}\mid{sinx}+{cosx}\mid+\:{x}\:−\int\frac{{sinx}}{{sinx}+{cosx}}{dx} \\ $$$$\Rightarrow\:_{\mathrm{0}} \int^{\frac{\pi}{\mathrm{4}}} \frac{{sinx}}{{sinx}+{cosx}}{dx}=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{x}−{ln}\mid{sinx}+{cosx}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\pi}{\mathrm{4}}\:−{ln}\sqrt{\mathrm{2}}\right)−\left(\mathrm{0}−{ln}\left(\mathrm{1}\right)\right)\right] \\ $$$$=\frac{\pi}{\mathrm{8}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\frac{\pi}{\mathrm{8}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{2} \\ $$ | ||