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what is the laplace transformation of cosx?

TEJKIRAT SINGH , 9 Years ago

Grade 12th pass

4 Answers

Vikas TU

Last Activity: 9 Years ago

= p/(p^2 +1)

Approved

Vikas TU

Last Activity: 9 Years ago

laplace is in cbse?

Drake

Last Activity: 9 Years ago

You need to integrate by parts twice, then you will find original integral in your integration result on right side. Solve resulting expression for this integral and plug in the limits of integration to obtain Laplace transform.

integrate by parts with
u=cos(b∙t) → du= -b∙sin(b∙t)
dv=e^(-s∙t) → v= -(1/s)∙e^(-s∙t)

∫ cos(b∙t)∙e^(-s∙t) dt
= -(1/s)∙e^(-s∙t)∙cos(b∙t) - (b/s)∙∫ e^(-s∙t)∙sin(b∙t) dt

to solve the integrals on right side integrate by parts with
u=sin(b∙t) → du= b∙cos(b∙t)
dv=e^(-s∙t) → v= -(1/s)∙e^(-s∙t)

∫ cos(b∙t)∙e^(-s∙t) dt
= -(1/s)∙e^(-s∙t)∙cos(b∙t) - (b/s)∙ [ -(1/s)∙e^(-s∙t)∙sin(b∙t) + (b/s)∙∫ e^(-s∙t)∙cos(b∙t) dt

∫ cos(b∙t)∙e^(-s∙t) dt
= -(1/s)∙e^(-s∙t)∙cos(b∙t) + (b/s²)∙e^(-s∙t)∙sin(b∙t) - (b²/s²)∙∫ e^(-s∙t)∙cos(b∙t) dt

solve for ∫ cos(b∙t)∙e^(-s∙t) dt
(1 + (b²/s²)) ∙ ∫ cos(b∙t)∙e^(-s∙t) dt
= -(1/s)∙e^(-s∙t)∙cos(b∙t) + (b/s²)∙e^(-s∙t)∙sin(b∙t)

((s² + b²)/s²) ∙ ∫ cos(b∙t)∙e^(-s∙t) dt
= -(1/s)∙e^(-s∙t)∙cos(b∙t) + (b/s²)∙e^(-s∙t)∙sin(b∙t)

∫ cos(b∙t)∙e^(-s∙t) dt
= [ s²/(s² + b²)] ∙( -(1/s)∙e^(-s∙t)∙cos(b∙t) + (b/s²)∙e^(-s∙t)∙sin(b∙t))
= (-s∙cos(b∙t) + b∙sin(b∙t))∙e^(-s∙t)/(s² + b²)

Hence:
ℒ{ cos(b∙t) }
...∞
= ∫ cos(b∙t)∙e^(-s∙t) dt
...0
= limt→∞{ (-s∙cos(b∙t) + b∙sin(b∙t))∙e^(-s∙t)/(s² + b²) }
- (-s∙cos(b∙0) + b∙sin(b∙0))∙e^(-s∙0)/(s² + b²)
because the exponential term vanishes for t→∞
= 0 - (-s∙1 + b∙0)∙1/(s² + b²)
= s/(s² + b²)

Hit approve if u got even a little help by my answer :)

Thanks .