How to solve the following integral?
∫π203√sin8xcos4xdx
Preferably without the universal substitution sin(t)=2tan(t/2)1+tan2(t/2)
Answer
Using B(a,b)=2∫π/20sin2a−1xcos2b−1xdx, your integral is12B(116,76)=Γ(116)Γ(76)2Γ(3)=5144Γ(56)Γ(16)=5π144cscπ6=5π72.Here the first = uses B(a,b)=Γ(a)Γ(b)Γ(a+b), the second Γ(a+1)=aΓ(a), the third Γ(a)Γ(1−a)=πcscπa.
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