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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
a2+b2
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B
(a2+b2)13
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C
a2+b2
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D
(a2b2)
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Solution

The correct option is C a2+b2
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). eaxaeaxa ( sin (bx)). b dx
Or I=cos(bx). eaxa+baeax ( 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). eaxaeaxa (cos (bx)). b dx
Or I1=sin(bx). eaxabaeax (cos (bx)). dx
Let's substitute I1=sin(bx). eaxabaeax (cos (bx)). dx in 1st equation.
I=cos(bx). eaxa+ba (sin(bx). eaxabaeax (cos (bx)). dx)
Or I=cos(bx). eaxa+ba2 sin(bx) eax.b2a2eax (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). eaxb2a2 (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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