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Question Number 137315 by liberty last updated on 01/Apr/21
∫0π/4cos3(2x)dx?
Answered by rs4089 last updated on 01/Apr/21
∫0π/4cos3(2x)dx2x=t⇒2dx=dt∫0π/212cos32(t).dt1212Γ(54)Γ(12)Γ(74)
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