Question

If $\int x{e}^{-5x^2}\sin 4x^{2}dx=e^{-5x^2}(A\sin 4x^2+B\cos 4x^2)+C$, then $A+B$

• A. $-\frac{1}{66}$
• B. $\frac{1}{66}$
• C. $-\frac{9}{66}$
• D. $-\frac{7}{66}$

Answer 1

Answer: (C)

$-\frac{9}{66}$

Given, $\int x{e}^{-5x^2}\sin 4x^{2}dx=K{e}^{-5x^2}(A\sin 4x^2+B\cos 4x^2)+C$.
Let $I=\int x{e}^{-5x^2}\sin 4x^{2}dx$.
Put $x^2=t$.
$\Rightarrow xdx=\frac{dt}{2}$
$\Rightarrow I=\frac{1}{2}\int e^{-5t}\sin 4tdt$
Using Integration by parts, we get
$=\frac{1}{2}[\sin 4t\frac{e^{-5t}}{-5}-\int 4\cos 4t\frac{e^{-5t}}{-5}dt]$
$=-\frac{1}{10}{e}^{-5t}\sin 4t+\frac{4}{10}\int e^{-5t}\cos 4tdt$
$=-\frac{1}{10}{e}^{-5t}\sin 4t+\frac{4}{10}[\cos 4t\frac{e^{-5t}}{-5}-\int (-4)\sin 4t\frac{e^{-5t}}{-5}.dt]$
$\Rightarrow I=-\frac{1}{10}{e}^{-5t}\sin 4t-\frac{2}{25}\cos 4t.e^{-5t}-\frac{8}{25}I+C$
$\Rightarrow \frac{33}{25}I=-\frac{1}{10}{e}^{-5t}\sin 4t-\frac{2}{25}\cos 4t.e^{-5t}+C$
$I=\frac{-5}{66}{e}^{-5x^2}\sin 4x^2-\frac{2}{33}{e}^{-5x^2}\cos 4x^2+C$
$I=e^{-5x^2}[-\frac{5}{66}\sin 4x^2-\frac{2}{33}\cos 4x^2]+C$
On comparing with given expression,
$A=-\frac{5}{66},B=-\frac{2}{33}$
$\therefore A+B=-\frac{9}{66}$