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Question

# If ∫eax cos (bx) dx=eaxK (a cosbx+b sinbx)+C, then the K here would be equal to -

A
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Solution

## The correct option is C We’ll use integration by parts method to calculate ∫eax cos (bx) dx being cos(bx) as the first function and eax being the second function. Let I=∫eax cos (bx) dx Or I=cos(bx). eaxa−∫eaxa (− sin (bx)). b dx Or I=cos(bx). eaxa+ba∫eax ( sin (bx)). dx Let I1=∫eax ( sin (bx)). dx I=cos(bx). eaxa+ba I1…(1) Applying integration by parts to integrate I1, sin(bx) being the first function, and eax being the second function. I1=sin(bx). eaxa−∫eaxa (cos (bx)). b dx Or I1=sin(bx). eaxa−ba∫eax (cos (bx)). dx Let's substitute I1=sin(bx). eaxa−ba∫eax (cos (bx)). dx in 1st equation. I=cos(bx). eaxa+ba (sin(bx). eaxa−ba∫eax (cos (bx)). dx) Or I=cos(bx). eaxa+ba2 sin(bx) eax.−b2a2∫eax (cos (bx)). dx We can see that in above equation we have ∫eax cos (bx) dx which is nothing but "I" itself. I=cos(bx). aaxa+ba2 sin(bx). eax−b2a2 (1) Or I (1+b2a2)=cos(bx). eaxa+ba2 sin(bx).eax Or I=a2a2+b2eax(acos(bx)+bsin(bx))a2 I=eaxa2+b2 (acos(bx)+b sin(bx)+C On comparing we can see that ‘K’ is equal to a2+b2.

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