calcilate-pi-6-pi-4-sin-x-cos-x-cos-2x-dx-

Question Number 40505 by prof Abdo imad last updated on 23/Jul/18

c a l c i l a t e \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{s i n (x)}{c o s (x) + c o s (2 x)} d x

Answered by MJS last updated on 23/Jul/18

\cos 2 x = {2 \cos}^{2} x - 1

\int \frac{\sin x}{{2 \cos}^{2} x + \cos x - 1} d x =

[t = \cos x \to d x = - \frac{d t}{\sin x}]

= - \int \frac{d t}{2 t^{2} + t - 1} = - \int \frac{d t}{(2 t - 1) (t + 1)} =

= \frac{1}{3} \int \frac{d t}{t + 1} - \frac{2}{3} \int \frac{d t}{2 t - 1} =

= \frac{1}{3} \ln (t + 1) - \frac{1}{3} \ln (2 t - 1) = \frac{1}{3} \ln \frac{t + 1}{2 t - 1} =

= \frac{1}{3} \ln ∣ \frac{\cos x + 1}{2 \cos x - 1} ∣ + C

\underset{\frac{π}{6}}{\int^{\frac{π}{4}}} \frac{\sin x}{\cos x + \cos 2 x} d x = \frac{1}{3} (\ln \frac{4 + 3 \sqrt{2}}{2} - \ln \frac{5 + 3 \sqrt{3}}{4}) \approx . 160153

Commented by math khazana by abdo last updated on 23/Jul/18

t h a n k y o u s i r M j s .