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

$$\underset{\:\mathrm{0}} {\overset{\:\pi/\mathrm{2}} {\int}}\frac{{cos}^{\mathrm{2}} {x}}{{cos}\left({x}−\pi/\mathrm{4}\right)}\:{dx} \\$$

Commented bymr W last updated on 27/Apr/21

$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}−\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$

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

$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}\mathrm{dx}...\left(\mathrm{1}\right),\:\mathrm{u}=\frac{\pi}{\mathrm{2}}−\mathrm{x} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)}\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}\mathrm{dx}...\left(\mathrm{2}\right) \\$$ $$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\mid\mathrm{sec}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\mid\mathrm{sec}\left(\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}\right)\mid−\mathrm{ln}\mid\mathrm{sec}\left(−\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\left(−\frac{\pi}{\mathrm{4}}\right)\mid\right] \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\mid\sqrt{\mathrm{2}}+\mathrm{1}\mid−\mathrm{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}\mid\right]=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\mid=\mathrm{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\$$

Commented bymathdanisur last updated on 30/Apr/21

$${thankyou}\:{sir} \\$$