Question

Salut, ma puteti ajuta la problema 457...Am tot incercat sa o scot la capat dar nu-mi iasa deloc, ma deranjeaza acel "x"...Aveti o idee?

Answer 1

$\displaystyle I = \int_{0}^{\frac{\pi}{4}} x\cdot \dfrac{\sin x}{\cos^3 x} \, dx \\ \\ \\ \int \dfrac{\sin x}{\cos^ 3x} \, dx = -\int (\cos x)'\cdot \cos^{-3}x \,dx =\\ \\ = -\dfrac{\cos^{-2}x}{-2}+ C = \dfrac{1}{2\cos^ 2 x} + C\\ \\ \\ I = \int_{0}^{\frac{\pi}{4}} x\cdot \Big(\dfrac{1}{2\cos^ 2 x}\Big)' \, dx = \\ \\ = \dfrac{x}{2\cos^2 x}\Big|_0^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} 1\cdot \dfrac{1}{2\cos^ 2 x} \, dx = \\ \\ =\dfrac{\frac{\pi}{4}}{2\cdot \frac{2}{4}} - \dfrac{\text{tg}\,x}{2} \Big|_0^{\frac{\pi}{4}} = \boxed{\dfrac{\pi}{4} - \dfrac{1}{2}}$

Answer 2

Ai poza mai jos........