Question

а) $\int\limits \frac{dx}{4-\frac{2x}{5} }$б) $\int\limits^\frac{\pi }{2} _0 e^{sinx} {cosx} dx$ CРОЧНО! ПОЖАЛУЙСТА!

Answer 1

$\int \frac{1}{4-\frac{2x}{5} }dx\\\\ZAMENA:t=4-\frac{2x}{5}\\\\\int-\frac{5}{2t}dt=-\frac{5}{2}\int\frac{1}{t}dt=-\frac{5}{2}*ln\mi(\mid t\mid)=-\frac{5}{2}*ln(\mid 4-\frac{2x}{5}\mid)+C\\\\\\\\\int^{\frac{\pi}{2} }_0e^{sin(x)}cos(x)dx\\\\ZAMENA:t=sin(x)\\\\\int e^tdt=e^t=e^{sin(x)}\mid ^{\frac{\pi}{2} }_0=e^{sin\frac{\pi}{2} }-e^{sin(0)}=e-1$