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# Calculus/Calculus 2 definite integral

Question
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

If the integral evaluates to the upper limit cubed,
we must have been integrating 3x².

To evaluate ∫f(cos(x))sin(x)dx given the f(x) = 3x², we get ∫3cos²(x)sin(x)dx.
Letting u(x) = cos(x) gives du = - sin(x)dx.
This means the problem is -∫3u²du which is = -u³.

To solve, put back in u(x) = cos(x) for u to get F(x), giving F(x) = -cos³(x).
Once this has been done, the answer would be F(π/2) - F(0).
Since it is known that cos(π/2) = 0 and cos(0) = 1, we have 0³ - 1³ = 0 - 1 = -1.

Calculus

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