If In=∫π01−sin2nx1−cos2xdx then I1,I2,I3,... are in
A
A.P
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B
G.P
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C
H.P
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D
None of these
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Solution
The correct option is A A.P In=∫π01−sin2nx1−cos2xdx Consider, In+In+2−2In+1=∫π01−sin2nx+(1−sin(2n+4)x)−2(1−sin(2n+2)x)1−cos2xdx =∫π02sin(2n+2)x−(sin(2n+4)x+sin2nx)1−cos2xdx =∫π02sin(2n+2)x−2sin(2n+2)xcos2x1−cos2xdx =2∫π0sin(2n+2)xdx =−2cos(2n+2)x2n+2π0 =−1n+1[cos(n+1)2π−cos0] =−1n+1(1−1)=0 ⇒2In+1=In+In+2 ∴In,In+1,In+2,.... are in A.P.