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Question

If π01a+bcosxdx=πa2b2, then π01(a+bcosx)2dx=?

(Note : a>0)

A
πa(a2b2)3/2
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B
πb(a2b2)3/2
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C
π(a2b2)3/2
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D
π
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Solution

The correct option is A πa(a2b2)3/2
Let f(a)=π01a+bcosxdx
On differentiating wrt. a we get,
f(a)=π01(a+bcosx)2dx
We have, f(a)=πa2b2
on differentiating above function w.r.t a we get,
Hence, f(a)=πa(a2b2)3/2

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