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

$$\mathrm{Prove}\mathrm{that}$$$${\int }_0^{\frac{\pi }{2}}\frac{\sin {}^2\mathrm{x}}{(\sin \mathrm{x}+\cos \mathrm{x})}\mathrm{dx}=\frac{1}{\sqrt{2}}\log (\sqrt{2}+1)$$

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

$$I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{\sin x+\cos x}dx...\mathrm{eqn}\left(i\right)$$$$I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{\sin x+\cos x}dx...\mathrm{eqn}\left(ii\right)[\mathrm{letting}t=\frac{\pi }{2}-x]$$$$\mathrm{eqn}\left(i\right)+\mathrm{eqn}\left(ii\right)\Rightarrow$$$$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x+\cos x}dx={\int }_0^{\frac{\pi }{2}}\frac{1}{\sin x+\cos x}dx$$$$I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1}{\sin x+\cos x}dx=\frac{1}{2}{\int }_0^1\frac{1}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}dt$$$$={\int }_0^1\frac{1}{1+2t-t^2}dt={\int }_0^1\frac{dt}{2-{(t-1)}^2}=\frac{1}{\sqrt{2}}{\left[\mathrm{argth}\left(\frac{t-1}{\sqrt{2}}\right)\right]}_0^1$$$$=\frac{1}{2\sqrt{2}}{[\ln \mid \frac{t+\sqrt{2}-1}{t-\sqrt{2}-1}\mid ]}_0^1=\frac{1}{2\sqrt{2}}\ln \mid \frac{\sqrt{2}+1}{\sqrt{2}-1}\mid =\frac{1}{2\sqrt{2}}\ln {(\sqrt{2}+1)}^2=\frac{1}{\sqrt{2}}\ln (\sqrt{2}+1)$$

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

$$\mathrm{thank}\mathrm{you}$$

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