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Question Number 168762 by Dildora last updated on 17/Apr/22

Commented by cortano1 last updated on 17/Apr/22
$∫ ((2sin (1/2)x cos (1/2)x)/(2cos (1/2)x(cos (1/2)x+sin (1/2)x))) dx = ∫ ((sin (1/2)x)/(cos (1/2)x+sin (1/2)x)) dx = ∫ ((2sin u)/(cos u+sin u)) du ; [(1/2)x=u ] ((2sin u)/(cos u+sin u)) = A(((sin u+cos u)/(sin u+cos u)))+B(((d(sin u+cos u))/(sin u+cos u))) ⇒2sin u= Asin u+Acos u+Bcos u−Bsin u ⇒2sin u=(A−B)sin u+(A+B)cos u { ((A−B=2)),((A+B=0)) :} ⇒ { ((A=1)),((B=−1)) :} I= ∫_0 ^( (π/4)) ((2sin u)/(sin u+cos u)) du = ∫_0 ^( (π/4)) du−∫_0 ^( (π/4)) ((d(sin u+cos u))/(sin u+cos u)) = (π/4)−[ ln ∣sin u+cos u∣ ]_( 0) ^(π/4) = (π/4)−ln (√2)$
$\int \frac{2 \sin \frac{1}{2} x \cos \frac{1}{2} x}{2 \cos \frac{1}{2} x (\cos \frac{1}{2} x + \sin \frac{1}{2} x)} d x$ $= \int \frac{\sin \frac{1}{2} x}{\cos \frac{1}{2} x + \sin \frac{1}{2} x} d x$ $= \int \frac{2 \sin u}{\cos u + \sin u} d u; [\frac{1}{2} x = u]$ $\frac{2 \sin u}{\cos u + \sin u} = A (\frac{\sin u + \cos u}{\sin u + \cos u}) + B (\frac{d (\sin u + \cos u)}{\sin u + \cos u})$ $\Rightarrow 2 \sin u = A \sin u + A \cos u + B \cos u - B \sin u$ $\Rightarrow 2 \sin u = (A - B) \sin u + (A + B) \cos u$ ${\begin{cases} A - B = 2 \\ A + B = 0 \end{cases} \Rightarrow {\begin{cases} A = 1 \\ B = - 1 \end{cases}$ $I = \int_{0}^{\frac{π}{4}} \frac{2 \sin u}{\sin u + \cos u} d u = \int_{0}^{\frac{π}{4}} d u - \int_{0}^{\frac{π}{4}} \frac{d (\sin u + \cos u)}{\sin u + \cos u}$ $= \frac{π}{4} - {[\ln ∣ \sin u + \cos u ∣]}_{0}^{\frac{π}{4}}$ $= \frac{π}{4} - \ln \sqrt{2}$
Commented by Dildora last updated on 17/Apr/22

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