Integrale con parametro

∫ ((2·k - 3)·SIN(x) + k·COS(x)) dx=

=(3 - 2·k)·COS(x) + k·SIN(x)

per x = 3/4·pi:

(3 - 2·k)·COS(3/4·pi) + k·SIN(3/4·pi)=

=(3 - 2·k)·(- √2/2) + k·(√2/2)=

=3·√2·k/2 - 3·√2/2

per x = pi/4:

(3 - 2·k)·COS(pi/4) + k·SIN(pi/4)=

=(3 - 2·k)·(√2/2) + k·(√2/2)=

=3·√2/2 - √2·k/2

Quindi:

3·√2·k/2 - 3·√2/2 - (3·√2/2 - √2·k/2) = 2·√2·k - 3·√2

Deve essere:

2·√2·k - 3·√2 = √2----> k = 2

$ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (2k-3) sin x + k cos x \, dx = \sqrt{2}$

$ \left. (3-2k) cos x + k sinx \right|_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} = \sqrt{2}$

$  -(3-2k)\frac{1}{\sqrt{2}}+ k\frac{1}{\sqrt{2}}- (3-2k)\frac{1}{\sqrt{2}} - k\frac{1}{\sqrt{2}} = \sqrt{2}$

$  -(3-2k)\frac{1}{\sqrt{2}}- (3-2k)\frac{1}{\sqrt{2}} = \sqrt{2}$

$  2(2k-3)\frac{1}{\sqrt{2}} = \sqrt{2}$

$  2(2k-3) =2$

$  2k-3 =1$

$ k = 2$