If u = int e^{ax} cos bx , dx and v = int e^{ | vedclass

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(C) Using integration by parts for $u = \int e^{ax} \cos bx \, dx$: $u = e^{ax} \frac{\sin bx}{b} - \int a e^{ax} \frac{\sin bx}{b} \, dx = \frac{e^{a

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$e^{2ax}$

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(C) Using integration by parts for $u = \int e^{ax} \cos bx \, dx$: $u = e^{ax} \frac{\sin bx}{b} - \int a e^{ax} \frac{\sin bx}{b} \, dx = \frac{e^{ax} \sin bx}{b} - \frac{a}{b}v$ Multiplying by $b$,we get $bu + av = e^{ax} \sin bx$ ... $(i)$ Similarly,for $v = \int e^{ax} \sin bx \, dx$: $v = e^{ax} \left( -\frac{\cos bx}{b} \right) - \int a e^{ax} \left( -\frac{\cos bx}{b} \right) \, dx = -\frac{e^{ax} \cos bx}{b} + \frac{a}{b}u$ Multiplying by $b$,we get $bv - au = -e^{ax} \cos bx$ ... $(ii)$ Squaring and adding equations $(i)$ and $(ii)$: $(bu + av)^2 + (bv - au)^2 = (e^{ax} \sin bx)^2 + (-e^{ax} \cos bx)^2$ $b^2u^2 + a^2v^2 + 2abuv + b^2v^2 + a^2u^2 - 2abuv = e^{2ax} (\sin^2 bx + \cos^2 bx)$ $(a^2 + b^2)u^2 + (a^2 + b^2)v^2 = e^{2ax}(1)$ $(a^2 + b^2)(u^2 + v^2) = e^{2ax}$

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) = $

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