Question

If $I_1={\int }_0^{\frac{\pi }{2}}{e}^{-x}sin4xdx$ and $I_2={\int }_0^{2\pi }{e}^{-x}sin4xdx$ and $I_2=\lambda I_1$, then $\lambda$ is equal to

• A. $\frac{e^{2\pi }-1}{e^{\pi }-1}$
• B. $\frac{e^{\pi }-1}{1-e^{2\pi }}$
• C. $\frac{1-e^{-2\pi }}{1-e^{-\frac{\pi }{2}}}$
• D. $\frac{1-e^{2\pi }}{1-e^{\frac{\pi }{2}}}$

Answer 1

The correct option is C $\frac{1-e^{-2\pi }}{1-e^{-\frac{\pi }{2}}}$

$I_1={\int }_0^{\frac{\pi }{2}}{e}^{-x}\sin 4xdx$
$I_2={\int }_0^{\frac{\pi }{2}}{e}^{-x}\sin 4xdx+{\int }_{\frac{\pi }{2}}^{\pi }{e}^{-x}\sin 4xdx+{\int }_{\pi }^{\frac{3\pi }{2}}{e}^{-x}\sin 4xdx+{\int }_{\frac{3\pi }{2}}^{2\pi }{e}^{-x}\sin 4xdx$
$I_3={\int }_{\frac{\pi }{2}}^{\pi }{e}^{-x}\sin 4xdx={\int }_0^{\frac{\pi }{2}}{e}^{-x-\frac{\pi }{2}}\sin (2\pi +4x)={\int }_0^{\frac{\pi }{2}}{e}^{-x}{e}^{-\frac{\pi }{2}}\sin (2\pi +4x)=e^{-\frac{\pi }{2}}{I}_1$
SO $I_2=(1+e^{-\frac{\pi }{2}}+e^{-\pi }+e^{-\frac{3\pi }{2}})I_1=\frac{1-e^{-2\pi }}{1-e^{-\frac{\pi }{2}}}{I}_1$