Question

Salut,am nevoie de ajutor la integrala de la punctul j)

Answer 1

$\displaystyle I =\int_{0}^{\frac{\pi}{4}} \dfrac{\sin x}{\cos x \sqrt{\cos 2x}}\, dx = \int_{0}^{\frac{\pi}{4}} \dfrac{\sin x}{\cos x \sqrt{\cos^2 x-\sin^2 x}}\, dx =\\ \\ = \int_{0}^{\frac{\pi}{4}} \dfrac{\sin x}{\cos x \sqrt{2\cos^2 x-1}}\, dx\\ \\ \text{Notez: }\cos x = t\Rightarrow -\sin x\, dx = dt\\ x = 0\Rightarrow t = 1\\ x = \dfrac{\pi}{4}\Rightarrow t = \dfrac{\sqrt{2}}{2}$

$\displaystyle I = -\int_{1}^{\frac{\sqrt{2}}{2}}\dfrac{1}{t\sqrt{2t^2-1}}\, dt=-\int_{1}^{\frac{\sqrt{2}}{2}}\dfrac{t}{t^2\sqrt{2t^2-1}}\, dt \\ \\\text{Notez: } \sqrt{2t^2-1} = y\Rightarrow 2t^2-1 = y^2 \\ \Rightarrow 4t\, dt = 2y\, dy \Rightarrow t\, dt = \dfrac{y}{2}\, dy\\ 2t^2 = y^2 + 1 \Rightarrow t^2 = \dfrac{y^2+1}{2}\\ t = 1 \Rightarrow y = 1\\ t = \dfrac{\sqrt{2}}{2} \Rightarrow y = 0$

$\displaystyle I = -\int_{1}^0 \dfrac{\dfrac{y}{2}}{\dfrac{y^2+1}{2}y}\, dy =\int_{0}^1 \dfrac{1}{y^2+1}\, dy = \\ = \text{arctg}\, y\Big|_{0}^1 = \text{arctg}\, 1 - \text{arctg}\, 0 = \boxed{\dfrac{\pi}{4}}$