If ∫(sinx)-112(cosx)-52dx=-p1q1(cotx)92-p2q2(cotx)52-p3q3(cotx)12+p4q4(cotx)-32+C where pi and qi are positive integers with gcd(pi,qi)=1 for i=1,2,3,4, and C is the constant of integration, then 15p1p2p3p4q1q2q3q4 is equal to ______ [2026]
(16)
∫(tanx)-11/2.sec8x dx
=∫(tanx)-11/2(1+tan2x)3sec2x dx
Put tanx=t
⇒∫t-11/2(1+t2)3 dt=∫t-11/2(1+3t2+3t4+t6) dt
=∫(t-11/2+3t-7/2+3t-3/2+t1/2)dt
=-29(cotx)9/2-65(cotx)5/2-6(cotx)1/2+23(cotx)-3/2+C
⇒p1=2, p2=6, p3=6, p4=2
and q1=9, q2=5, q3=1, q4=3
15p1p2p3p4q1q2q3q4=15·2·6·6·29·5·1·3=16