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Question Number 177298 by peter frank last updated on 03/Oct/22

Prove that

\int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{(\sin x + \cos x)} dx = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)

Answered by Ar Brandon last updated on 11/Oct/22

I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\sin x + \cos x} d x \dots eqn (i)

I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\sin x + \cos x} d x \dots eqn (i i) [letting t = \frac{π}{2} - x]

eqn (i) + eqn (i i) \Rightarrow

2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x + \cos^{2} x}{\sin x + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{1}{\sin x + \cos x} d x

I = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{1}{\sin x + \cos x} d x = \frac{1}{2} \int_{0}^{1} \frac{1}{\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \cdot \frac{2}{1 + t^{2}} d t

= \int_{0}^{1} \frac{1}{1 + 2 t - t^{2}} d t = \int_{0}^{1} \frac{d t}{2 - {(t - 1)}^{2}} = \frac{1}{\sqrt{2}} {[argth (\frac{t - 1}{\sqrt{2}})]}_{0}^{1}

= \frac{1}{2 \sqrt{2}} {[\ln ∣ \frac{t + \sqrt{2} - 1}{t - \sqrt{2} - 1} ∣]}_{0}^{1} = \frac{1}{2 \sqrt{2}} \ln ∣ \frac{\sqrt{2} + 1}{\sqrt{2} - 1} ∣= \frac{1}{2 \sqrt{2}} \ln {(\sqrt{2} + 1)}^{2} = \frac{1}{\sqrt{2}} \ln (\sqrt{2} + 1)

Commented by peter frank last updated on 03/Oct/22

thank you