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How do you find the integral of #(6t)/sqrt(36-t^2)#?

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Jun 2, 2015

Consider 36 -#t^2# =u, so that tdt=#-1/2# du

#int (6tdt)/sqrt(36-t^2)=int -6/2 (du)/sqrtu# = -3*2 #u^(1/2)#

= -6#sqrt(36-t^2)# +C

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