MCQ
If $u = \int_{}^{} {{e^{ax}}\cos bx\;dx} $ and $v = \int_{}^{} {{e^{ax}}\sin bx\;dx} $, then $({a^2} + {b^2})({u^2} + {v^2}) = $
  • A
    $2{e^{ax}}$
  • B
    $({a^2} + {b^2}){e^{2ax}}$
  • ${e^{2ax}}$
  • D
    $({a^2} - {b^2}){e^{2ax}}$

Answer

Correct option: C.
${e^{2ax}}$
c
(c)$u = \int_{}^{} {{e^{ax}}\cos bx\,dx} $$ = {e^{ax}}\frac{{\sin bx}}{b} - \frac{a}{b}\int_{}^{} {{e^{ax}}.\sin bx\,dx} $
$ = \frac{{{e^{ax}}\sin bx}}{b} - \frac{a}{b}v$ $ \Rightarrow bu + av = {e^{ax}}\sin bx$..$(i)$
Similarly $bv - au = - {e^{ax}}\cos bx$..$(ii)$
Squaring $(i)$ and $(ii)$ and adding, we get
$({a^2} + {b^2})({u^2} + {v^2}) = {e^{2ax}}$.

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