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Q.

In=0π/2xn.cosxdx then the value of I9+72I7=

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a

π29

b

9.π28

c

π28

d

0

answer is A.

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Detailed Solution

In=0π/2 xn. cos x dx    =xn(Sm x)]0π/2-0π/2n.xn-1(Sm x) dx    =π2n-n 0π/2xn-1.Sm x dx    =π2n-nxn-1.-(cos x)0π/2-0π/2(n-1)xn-2(-(cos x)dx)    =π2n-n0+(n-1) 0π/2xn-2 cos x dx    =π2n-n(n-1) In-2      In+n(n-1) In-2=π2n n=9  I9+72  I7==π29

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