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0-2-sin-x-sin-x-cos-x-dx-




Question Number 189874 by mathocean1 last updated on 23/Mar/23
∫_0 ^( 2)    ((sin(x))/(sin(x)+cos(x)))dx= ?
$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\:\:\frac{{sin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}=\:? \\ $$
Answered by ARUNG_Brandon_MBU last updated on 23/Mar/23
I=∫_0 ^2 ((sinx)/(sinx+cosx))dx=(1/( (√2)))∫_0 ^2 ((sin((x−(π/4))+(π/4)))/(cos(x−(π/4))))dx     =(1/( (√2)))∫_0 ^2 ((sin(x−(π/4))cos(π/4)+cos(x−(π/4))sin(π/4))/(cos(x−(π/4))))dx     =(1/( 2))[lncos(x−(π/4))]_2 ^0 +[(x/( 2))]_0 ^2      =(1/( 2))ln((1/( (√2))))−(1/( 2))ln(((cos2+sin2)/( (√2))))+1     =1−(1/( 2))ln(cos(2)+sin(2))
$${I}=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{sin}{x}}{\mathrm{sin}{x}+\mathrm{cos}{x}}{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{sin}\left(\left({x}−\frac{\pi}{\mathrm{4}}\right)+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{sin}\left({x}−\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\frac{\pi}{\mathrm{4}}+\mathrm{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}\frac{\pi}{\mathrm{4}}}{\mathrm{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\mathrm{2}}\left[\mathrm{lncos}\left({x}−\frac{\pi}{\mathrm{4}}\right)\right]_{\mathrm{2}} ^{\mathrm{0}} +\left[\frac{{x}}{\:\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\:\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{cos2}+\mathrm{sin2}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{1} \\ $$$$\:\:\:=\mathrm{1}−\frac{\mathrm{1}}{\:\mathrm{2}}\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{2}\right)+\mathrm{sin}\left(\mathrm{2}\right)\right) \\ $$
Commented by mathocean1 last updated on 26/Mar/23
Thanks...
$${Thanks}… \\ $$
Answered by mehdee42 last updated on 23/Mar/23
(1/2)∫_0 ^2 ((2sinx(cosx−sinx))/(cos^2 x−sin^2 x))dx=(1/2)∫_0 ^2 ((sin2x+cos2x−1)/(cos2x))dx  (1/2)∫_0 ^2 (tan2x+1−sec2x)dx=(1/2)(−(1/2)lncos2x+x−(1/2)ln(((cosx+sinx)/(cosx−sinx))))∣_0 ^2   (1/2)(−(1/2)lncos4+1−(1/2)ln(((cos2+sin2)/(cos2−sin2)))  =−(1/4)ln(cos2+sin2)−(1/4)ln(cos2+sin2)+1−(1/4)ln(cos2+sin2)+(1/4)ln(cos2−sin2)  =1−(1/2)ln(cos2+sin2)
$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{2}{sinx}\left({cosx}−{sinx}\right)}{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}−\mathrm{1}}{{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left({tan}\mathrm{2}{x}+\mathrm{1}−{sec}\mathrm{2}{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{lncos}\mathrm{2}{x}+{x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{cosx}+{sinx}}{{cosx}−{sinx}}\right)\right)\underset{\mathrm{0}} {\overset{\mathrm{2}} {\mid}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{lncos}\mathrm{4}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{cos}\mathrm{2}+{sin}\mathrm{2}}{{cos}\mathrm{2}−{sin}\mathrm{2}}\right)\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({cos}\mathrm{2}+{sin}\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({cos}\mathrm{2}+{sin}\mathrm{2}\right)+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({cos}\mathrm{2}+{sin}\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({cos}\mathrm{2}−{sin}\mathrm{2}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({cos}\mathrm{2}+{sin}\mathrm{2}\right) \\ $$
Commented by mathocean1 last updated on 26/Mar/23
Thanks...
$${Thanks}… \\ $$

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