Question

${\int }_0^{\pi /2}{e}^{-2x}co{s}^{2}xdx=$

• A. $\frac{3-e^{-\pi }}{8}$
• B. $\frac{e^{-\pi }-3}{8}$
• C. $\frac{3-e^{-\pi /2}}{8}$
• D. $\frac{e^{-\pi /2}-3}{8}$

Answer 1

The correct option is A $\frac{3-e^{-\pi }}{8}$

$I={\int }_0^{\pi /2}{e}^{-2x}co{s}^{2}xdx$
$={\int }_0^{\pi /2}\frac{e^{-2x}}{2}(1+cos2x)dx$
$={\int }_0^{\pi /2}\frac{e^{-2x}}{2}dx+{\int }_0^{\pi /2}\frac{e^{-2x}}{2}cos2xdx$
$={(-\frac{e^{-2x}}{4})}_0^{\frac{\pi }{2}}+{\left\{\frac{e^{-2x}}{\left(\right(-2{)}^2+2^2)2}\right(-2cos2x+2sin2x\left)\right\}}_0^{\pi /2}$
$=(-\frac{e^{-\pi }}{4}+\frac{1}{4})+\frac{e^{-\pi }}{16}\left(2\right)-\frac{1}{16}(-2)$
$=-\frac{e^{-\pi }}{8}+\frac{3}{8}=\frac{(3-e^{-\pi })}{8}$