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Question Number 139414 by mathdanisur last updated on 26/Apr/21

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{{cos}^{2}x}{cos(x-\pi /4)}dx$$

Commented by mr W last updated on 27/Apr/21

$$\frac{1}{2}{\left[\ln \frac{1+\sin (x-\frac{\pi }{4})}{1-\sin (x-\frac{\pi }{4})}\right]}_0^{\frac{\pi }{2}}=\frac{1}{2}\ln \frac{1+\frac{\sqrt{2}}{2}}{1-\frac{\sqrt{2}}{2}}=\ln (1+\sqrt{2})$$

Answered by Ar Brandon last updated on 27/Apr/21

$$={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^2\mathrm{x}}{\cos (\mathrm{x}-\frac{\pi }{4})}\mathrm{dx}...\left(1\right),\mathrm{u}=\frac{\pi }{2}-\mathrm{x}$$$$={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\mathrm{x}}{\cos (\frac{\pi }{4}-\mathrm{x})}\mathrm{dx}={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\mathrm{x}}{\cos (\mathrm{x}-\frac{\pi }{4})}\mathrm{dx}...\left(2\right)$$$$\left(1\right)+\left(2\right)$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}}{\cos (\mathrm{x}-\frac{\pi }{4})}\mathrm{dx}=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\cos (\mathrm{x}-\frac{\pi }{4})}$$$$=\frac{1}{2}{[\ln \mid \sec (\mathrm{x}-\frac{\pi }{4})+\tan (\mathrm{x}-\frac{\pi }{4})\mid ]}_0^{\frac{\pi }{2}}$$$$=\frac{1}{2}[\ln \mid \sec \left(\frac{\pi }{4}\right)+\tan \left(\frac{\pi }{4}\right)\mid -\ln \mid \sec (-\frac{\pi }{4})+\tan (-\frac{\pi }{4})\mid ]$$$$=\frac{1}{2}[\ln \mid \sqrt{2}+1\mid -\ln \mid \sqrt{2}-1\mid ]=\frac{1}{2}\ln \mid \frac{\sqrt{2}+1}{\sqrt{2}-1}\mid =\ln (\sqrt{2}+1)$$

Commented by mathdanisur last updated on 30/Apr/21

$$thankyousir$$

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