Question Number 168762 by Dildora last updated on 17/Apr/22
Commented by cortano1 last updated on 17/Apr/22
![∫ ((2sin (1/2)x cos (1/2)x)/(2cos (1/2)x(cos (1/2)x+sin (1/2)x))) dx = ∫ ((sin (1/2)x)/(cos (1/2)x+sin (1/2)x)) dx = ∫ ((2sin u)/(cos u+sin u)) du ; [(1/2)x=u ] ((2sin u)/(cos u+sin u)) = A(((sin u+cos u)/(sin u+cos u)))+B(((d(sin u+cos u))/(sin u+cos u))) ⇒2sin u= Asin u+Acos u+Bcos u−Bsin u ⇒2sin u=(A−B)sin u+(A+B)cos u { ((A−B=2)),((A+B=0)) :} ⇒ { ((A=1)),((B=−1)) :} I= ∫_0 ^( (π/4)) ((2sin u)/(sin u+cos u)) du = ∫_0 ^( (π/4)) du−∫_0 ^( (π/4)) ((d(sin u+cos u))/(sin u+cos u)) = (π/4)−[ ln ∣sin u+cos u∣ ]_( 0) ^(π/4) = (π/4)−ln (√2)](https://www.tinkutara.com/question/Q168763.png)
$$\:\int\:\frac{\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{2cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)}\:{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{2sin}\:{u}}{\mathrm{cos}\:{u}+\mathrm{sin}\:{u}}\:{du}\:;\:\left[\frac{\mathrm{1}}{\mathrm{2}}{x}={u}\:\right] \\ $$$$\frac{\mathrm{2sin}\:{u}}{\mathrm{cos}\:{u}+\mathrm{sin}\:{u}}\:=\:{A}\left(\frac{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}\right)+{B}\left(\frac{{d}\left(\mathrm{sin}\:{u}+\mathrm{cos}\:{u}\right)}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}\right) \\ $$$$\Rightarrow\mathrm{2sin}\:{u}=\:{A}\mathrm{sin}\:{u}+{A}\mathrm{cos}\:{u}+{B}\mathrm{cos}\:\:{u}−{B}\mathrm{sin}\:{u} \\ $$$$\Rightarrow\mathrm{2sin}\:{u}=\left({A}−{B}\right)\mathrm{sin}\:{u}+\left({A}+{B}\right)\mathrm{cos}\:{u} \\ $$$$\:\begin{cases}{{A}−{B}=\mathrm{2}}\\{{A}+{B}=\mathrm{0}}\end{cases}\:\Rightarrow\begin{cases}{{A}=\mathrm{1}}\\{{B}=−\mathrm{1}}\end{cases}\: \\ $$$$\:{I}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2sin}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}\:{du}\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {du}−\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{d}\left(\mathrm{sin}\:{u}+\mathrm{cos}\:{u}\right)}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}} \\ $$$$\:=\:\frac{\pi}{\mathrm{4}}−\left[\:\mathrm{ln}\:\mid\mathrm{sin}\:{u}+\mathrm{cos}\:{u}\mid\:\right]_{\:\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\:=\:\frac{\pi}{\mathrm{4}}−\mathrm{ln}\:\sqrt{\mathrm{2}} \\ $$
Commented by Dildora last updated on 17/Apr/22
$${thank}\:{you} \\ $$