Calculus/Calculus 2

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Question
I dont understand what to do for this problem.

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

Answer
Change variables , let u = cos(x) , du/dx = -sin(x) , dx = -du/sin(x)

S f(cos(x))sin(x)dx  = -S f(u) sin(x) du/sin(x) = -S f(u) du

Since u = cos (x) , and the limits for x are from 0 to pi/2 ,  the limits on u will be from
cos(0) to cos(pi/2) , which will be from 1 to 0

We want -S f(u) du from 1 to 0 , which is S f(u) du from 0 to 1 .

We are told

0 to a S f(t)dt = a^3

so 0 to 1 S f(t)dt = 1^3 = 1

The answer is 1

Calculus

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I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.

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