Question

${\int }_0^{\infty }{e}^{-2x}(Cos4x+Sin4x)dx=$

• A. $4/25$
• B. $3/10$
• C. $1/5$
• D. $7/25$

Answer 1

The correct option is B $3/10$

$I={\int }_0^{\infty }{e}^{-2x}(cos4x+sin4x)dx={\int }_0^{\infty }{e}^{-2x}sin4xdx+{\int }_0^{\infty }{e}^{-2x}cos4xdx$
Using $\int e^{ax}sinbxdx=\frac{e^{ax}}{a^2+b^2}(asinbx-bcosbx)+c\int e^{ax}cosbxdx=\frac{e^{ax}}{a^2+b^2}(acosbx+bsinbx)+c$
$I={\left[\frac{e^{-2x}}{4+16}(-2sin4x-4cos4x)\right]}_0^{\infty }+{\left[\frac{e^{-2x}}{4+16}(-2cos4x+sin4x)\right]}_0^{\infty }=\frac{4}{20}+\frac{2}{20}=\frac{3}{10}$