## Step 1: Setup and Integration by Parts We want to find the Laplace transform of \(\cos(at)\): \[ \{\cos(at)\}(s) = \int_{}^{\infty} e^{-st} \cos(at)\,dt \] Let: - \( u = \cos(at) \) - \( dv = e^{-st}dt \) - Hence, \( du = -a\sin(at)dt \), \( v = -\frac{1}{s} e^{-st} \) Apply integration by parts: \[ \int u\,dv = uv - \int v\,du \] \[ = \left[ -\frac{1}{s} e^{-st} \cos(at) \right]_^\infty - \int_^\infty \left(-\frac{1}{s} e^{-st}\right)(-a\sin(at))\,dt \] \[ = \left[ -\frac{1}{s} e^{-st} \cos(at) \right]_^\infty - \frac{a}{s} \int_^\infty e^{-st} \sin(at)\,dt \] **Explanation:** We use integration by parts to split the original integral into two terms: one evaluated at the limits and another involving the Laplace transform of \(\sin(at)\). --- ## Step 2: Integrate \(\sin(at)\) Term and Solve for the Original Integral Now, let’s compute \(\int_^\infty e^{-st} \sin(at)\,dt\) using integration by parts again: Let: - \( u = \sin(at) \) - \( dv = e^{-st}dt \) - Hence, \( du = a\cos(at)dt \), \( v = -\frac{1}{s} e^{-st} \) \[ \int_^\infty e^{-st} \sin(at)\,dt = \left[ -\frac{1}{s} e^{-st} \sin(at) \right]_^\infty + \frac{a}{s} \int_^\infty e^{-st} \cos(at)\,dt \] Notice, the last integral is the same as the one we started with! Let \( I = \int_^\infty e^{-st} \cos(at)\,dt \): Substitute back and collect like terms: \[ I = \left[ -\frac{1}{s} e^{-st} \cos(at) \right]_^\infty - \frac{a}{s} \left( \left[ -\frac{1}{s} e^{-st} \sin(at) \right]_^\infty + \frac{a}{s}I \right ) \] The terms at infinity vanish due to the exponential decay, and at \( t= \), \(\cos()=1\), \(\sin()=\): \[ I = - \left(-\frac{1}{s}\right) - \frac{a}{s} \left( + \frac{a}{s}I \right ) \] \[ I = \frac{1}{s} - \frac{a^2}{s^2}I \] \[ I \left(1 + \frac{a^2}{s^2}\right) = \frac{1}{s} \] \[ I = \frac{1}{s(1 + \frac{a^2}{s^2})} = \frac{1}{s} \cdot \frac{s^2}{s^2 + a^2} \] \[ I = \frac{s}{s^2 + a^2} \] --- ## **Final Answer** \[ \boxed{ \int_{}^{\infty} e^{-st} \cos(at)\,dt = \frac{s}{s^2 + a^2} } \] **Explanation:** By applying integration by parts twice and recognizing the recurrence, we solve for the original Laplace transform integral and find the standard result.