## S_Student 2 years ago laplace cosat and sinat

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1. nitz

L(cos at)=$s/(s ^{2}+a ^{2})$ and L(sin at)=$a/(s ^{2}+a ^{2})$

2. nitz

if u are interested i can tell u the derivation for these

3. S_Student

thanx friend...and ya i want to expain me in detail..

4. nitz

ok ...

5. S_Student

thnx

6. nitz

$e ^{iat}=\cos(at)+isin(at)$ taking laplace on both sides:: $L(e ^{iat})=L(\cos (at))+iL(\sin at)$...............(1) we know that :$L(e ^{iat})=1/(s-ia)$ now rationalizing(ie multiplying numerator and denominator by (s+ia): $L(e ^{iat})=(s+ia)/((s+ia)(s-ia))$ $L(e ^{iat})=(s+ia)/(s ^{2}+a ^{2})$ $L(e ^{iat})=(s/(s ^{2}+a ^{2}))+i (a/(s ^{2}+a ^{2}))$................(2) compare 1 and 2::

7. nitz

U GET : $L(\cos at)=s/(s ^{2}+a ^{2})$ and $L(\sin at)=a/(s ^{2}+a ^{2})$

8. S_Student

it is enough for me..

9. S_Student

thanx dear..

10. UnkleRhaukus

Slight variation ____________ \begin{align*} \mathcal L\left\{ \sin({n}t)\right\}(p)&=\int\limits_0^\infty \sin({n}t)e^{-{p}t}\text dt\\ \\&=\int\limits_0^\infty {\frak I}\left(e^{i{n}t}\right)e^{-{p}t}\text dt\\ \\&={\frak I}\left(\int\limits_0^\infty e^{i{n}t}e^{-{p}t}\text dt\right)\\ \\&={\frak I}\left(\int\limits_0^\infty e^{-({p}-i{n})t}\text dt\right)\\ \\&={\frak I}\left(\left.\frac{e^{-({p}-i{n})t}}{-({p}-i{n})}\right|_0^\infty\right)\\ \\&={\frak I}\left(\frac{1}{{p}-i{n}}\right)\\ \\&={\frak I}\left(\frac{1}{{p}-i{n}}\times\frac{{p}+i{n}}{{p}+i{n}}\right)\\ \\&={\frak I}\left(\frac{{p}+i{n}}{{p}^2+{n}^2}\right)\\ \\&= \frac{n}{{p}^2+{n}^2}\\ \end{align*} ____________ \begin{align*} \mathcal L\left\{\cos ({n}t)\right\}(p)&=\int\limits_0^\infty \cos({n}t)e^{-{p}t}\text dt\\ \\&=\int\limits_0^\infty {\frak R}\left(e^{i{n}t}\right)e^{-{p}t}\text dt\\ \\&={\frak R}\left(\int\limits_0^\infty e^{i{n}t}e^{-{p}t}\text dt\right)\\ \\&={\frak R}\left(\int\limits_0^\infty e^{-({p}-i{n})t}\text dt\right)\\ \\&={\frak R}\left(\left.\frac{e^{-({p}-i{n})t}}{-({p}-i{n})}\right|_0^\infty\right)\\ \\&={\frak R}\left(\frac{1}{{p}-i{n}}\right)\\ \\&={\frak R}\left(\frac{1}{{p}-i{n}}\times\frac{{p}+i{n}}{{p}+i{n}}\right)\\ \\&={\frak R}\left(\frac{{p}+i{n}}{{p}^2+{n}^2}\right)\\ \\&= \frac{{p}}{{p}^2+{n}^2}\\ \end{align*}