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Prove-0-1-K-2-36K-2-6K-K-2-36K-2-dk-k-1-k-2-2-3-pi-2-4-2-2-3-




Question Number 217683 by MrGaster last updated on 18/Mar/25
Prove:∫_0 ^1 (√((((√(K^2 +36K′^2 ))+6K^′ )/(K^2 +36K^(′2) )) ))(dk/( (√k)(1−k^2 )^(2/3) ))=(√π)((√2)−(√((4−2(√2))/3)))
Prove:01K2+36K2+6KK2+36K2dkk(1k2)23=π(24223)

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