Say that integrate from 0 to a of f(t)dt = a^3.  Evaluate integrate from 0 to pi/2 of f(cos(x))sin(x)dx  

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Since the problem provides the information that `int_0^a f(t)dt = a^3` , hence, reasoning by analogy, you may consider `cos x = t => -sin x dx = dt ` such that:

`int_0^(pi/2) f(cos x) sin x dx = -int_0^1f(t)(-dt) = int_0^1 f(t)dt`

Notice that you need to change the limits of integration such that:

`x = 0 => cos 0 = 1`

`x = pi/2 => cos(pi/2) = 0`

`int_1^0 f(t)dt = -int_0^1 f(t) dt`

Considering upper limit of integration a = 1 and using the condition provided by the problem yields:

`int_0^1 f(t)dt = 1^3 => int_0^1 f(t)dt = 1`

Hence, evaluating the given definite integral, using the conditions provided by the problem yields `int_0^(pi/2) f(cos x) sin x dx = 1.`

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