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what is the laplace transformation of cosx?

what is the laplace transformation of cosx?

Grade:12th pass

4 Answers

Vikas TU
14149 Points
6 years ago
= p/(p^2 +1)
 
 
Vikas TU
14149 Points
6 years ago
laplace is in cbse?
Drake
277 Points
6 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 . 
VAISHNAVI BATHINA
130 Points
6 years ago
S/(S^2+1)

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