Given,
bn=∫02πsinx1+cos2nxdx
bn+1−bn=∫02πsinxcos2(n+1)x−cos2nxdx
=∫02πsinx−sin(2n+1)xsinxdx
=(2n+1cos(2n+1)x)02π=2n+1−1
So,b3−b21=−5
b4−b31=−7
b5−b41=−9
So,b3−b21,b4−b31,b5−b41 are in A.P. with c.d=−2
JEE Main 2022 — Mathematics Calculus
If bn=∫02πsinxcos2nxdx,n∈N, then
Held on 25 Jun 2022 · Verified 6 Jul 2026.
b3−b2,b4−b3,b5−b4 are in an A.P. with common difference−2
b3−b21,b4−b31,b5−b41 are in an A.P. with common difference 2
b3−b2,b4−b3,b5−b4 are in a G.P.
b3−b21,b4−b31,b5−b41 are in an A.P. with common difference −2
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