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$({a^2} - {b^2}){e^{2ax}}$
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