∫0π/2 Sin2⁡xSin⁡x+Cos⁡xdx=

I+I=∫0π/2 cos2⁡xsin⁡x+cos⁡xdx+sin2⁡xsin⁡x+cos⁡xdx ⇒2I=∫0π21sinx+cosxdx =∫0π2cosce(x+π4)dx =12log(2+1)

∫0π/2 Sin2⁡xSin⁡x+Cos⁡xdx=

I+I=∫0π/2 cos2⁡xsin⁡x+cos⁡xdx+sin2⁡xsin⁡x+cos⁡xdx ⇒2I=∫0π21sinx+cosxdx =∫0π2cosce(x+π4)dx =12log(2+1)

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$\int_{0}^{π / 2} \frac{{Sin}^{2} x}{Sin x + Cos x} dx =$

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$\frac{1}{\sqrt{2}} \ln (\sqrt{2} + 1)$

$n (\sqrt{2} + 1)$

$\sqrt{2} \ln (\sqrt{2} + 1)$

answer is C.

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Detailed Solution

$I + I = \int_{0}^{π / 2} \frac{\cos^{2} x}{\sin x + \cos x} dx + \frac{\sin^{2} x}{\sin x + \cos x} dx \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \frac{1}{sinx + cosx} dx = \int_{0}^{\frac{π}{2}} cosce (x + \frac{π}{4}) dx = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$