Question

If $\int x{e}^{-5x^2}sin4x^{2}dx=K{e}^{-5x^2}(Asin4x^2+Bcos4x^2)+C$. Then

• A. $\frac{1}{82},5,4$
• B. $\frac{1}{82},-5,4$
• C. $\frac{1}{82},-5,-4$
• D. $\frac{1}{82},\frac{-1}{5},\frac{-1}{4}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{82},5,4$
C $\frac{1}{82},-5,-4$

let , $I=\int x{e}^{-5x^2}\sin \left(4x^2\right)dx$
put $t=x^2\Rightarrow dt=2xdx$
$I=\frac{1}{2}\int e^{-5t}\sin \left(4t\right)dt$
let , $K=\int e^{-5t}\sin \left(4t\right)dt$
using integration by parts,
$=e^{-5t}\int \sin \left(4t\right)dt-\int (-5e^{-5t})\int \sin \left(4t\right)dt$
$=-e^{-5t}\frac{\cos \left(4t\right)}{4}-\frac{5}{4}\int e^{-5t}\cos \left(4t\right)dt$
using integration by parts again,
$=-e^{-5t}\frac{\cos \left(4t\right)}{4}-\frac{5}{4}[e^{-5t}\frac{\sin \left(4t\right)}{4}+\frac{5}{4}\int e^{-5t}\sin \left(4t\right)dt]$
$K=-e^{-5t}\frac{\cos \left(4t\right)}{4}-\frac{5}{16}{e}^{-5t}\sin \left(4t\right)-\frac{25}{16}K$
$K=-\frac{1}{41}{e}^{-5t}\left(4\cos \left(4t\right)+5\sin \left(4t\right)\right)$
$K=-\frac{1}{41}{e}^{-5t}\left(4\cos \left(4t\right)+5\sin \left(4t\right)\right)$
substituting $t=x^2$
$I=\frac{1}{2}K=-\frac{1}{82}{e}^{-5x^2}\left(4\cos \left(4x^2\right)+5\sin \left(4x^2\right)\right)$
Threrefore , $A=5$ $K=-\frac{1}{82}$