Question

Алгебра..........................

Answer 1

Ответ:

$\int ^{ \frac{\pi}{3} }_{0} {e}^{cosx} sinxdx = |t = cosx;dt = - sinxdx;cos0 = 1;cos \frac{\pi}{3} = \frac{1}{2}| = \\ = - \int ^{ \frac{1}{2} } _{1} {e}^{t} dt = \int^{1}_{ \frac{1}{2} } {e}^{t} dt = {e}^{t} | ^{ 1 }_{ \frac{1}{2} } = e - \sqrt{e}$

$\int ^{1} _{ - 1} \frac{dx}{ {x}^{2} + 2x + 5 } = \int ^{1} _{ - 1} \frac{dx}{ {x}^{2} + 2x + 1 + 4 } = \\ = \int ^{1} _{ - 1} \frac{dx}{{(x + 1 )}^{2} + 4 } = \frac{1}{2} arctg \frac{x + 1}{2} | ^{1} _{ - 1} = \frac{1}{2} (arctg1 - arctg0) = \frac{\pi}{8}$