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Question Number 200934 by sonukgindia last updated on 26/Nov/23

Answered by Frix last updated on 27/Nov/23

∫_0 ^(1/2) (tan πx)^(4/5) dx =^(t=(cot x)^(1/5) )   =−(5/π)∫_∞ ^0  (dt/(t^(10) +1)) =(5/π)∫_0 ^∞ (dt/(t^(10) +1))=((1+(√5))/2)

120(tanπx)45dx=t=(cotx)15=5π0dtt10+1=5π0dtt10+1=1+52

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