0-pi-2-cos-2-x-cos-x-pi-4-dx-

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

\underset{0}{\int^{π / 2}} \frac{{c o s}^{2} x}{c o s (x - π / 4)} d x

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

\frac{1}{2} {[\ln \frac{1 + \sin (x - \frac{π}{4})}{1 - \sin (x - \frac{π}{4})}]}_{0}^{\frac{π}{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{π}{2}} \frac{\cos^{2} x}{\cos (x - \frac{π}{4})} dx \dots (1), u = \frac{π}{2} - x

= \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\cos (\frac{π}{4} - x)} dx = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\cos (x - \frac{π}{4})} dx \dots (2)

(1) + (2)

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x + \sin^{2} x}{\cos (x - \frac{π}{4})} dx = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{dx}{\cos (x - \frac{π}{4})}

= \frac{1}{2} {[\ln ∣ \sec (x - \frac{π}{4}) + \tan (x - \frac{π}{4}) ∣]}_{0}^{\frac{π}{2}}

= \frac{1}{2} [\ln ∣ \sec (\frac{π}{4}) + \tan (\frac{π}{4}) ∣ - \ln ∣ \sec (- \frac{π}{4}) + \tan (- \frac{π}{4}) ∣]

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

Commented by mathdanisur last updated on 30/Apr/21

t h a n k y o u s i r