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Use trigonometric substitution to verify that

∫0xa2−t2−−−−−−√dt
〖√(a^2-t^2 ) dt〗=1/2 a^2 sin^(-1)⁡〖x/a〗+1/2 x√(a^2-x^2 )


Use the figure to give trigonometric interpretations of both terms on the right side of the equation in part (a).

1 Answer

3 votes
t = 0
=> Q = 0
t = x
=> Q = arcsin(x/a)
rest it just simple integration of trigonometric function
User Rafael Neto
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