∫sin−11/2xcos−5/2xdx.
Since m+n=−8 (negative even), substitute t=cotx, dx=1+t2−dt.
Integral becomes −∫t5/2(1+t2)3dt=−∫(t−5/2+3t−1/2+3t3/2+t7/2)dt.
=32t−3/2−6t1/2−56t5/2−92t9/2+C.
Comparing: q1p1=92, q2p2=56, q3p3=16, q4p4=32.
q1q2q3q415p1p2p3p4=9⋅5⋅1⋅315⋅2⋅6⋅6⋅2=1352160=16.