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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

I am not really sure how to tackle this, but nevertheless here is my attempt;If we let μ=sin(x) then dμ/dx=cosx→dμ=cos(x)dx.That means that ∫π/20f(sin(x))dx=∫00f(μ)cosxdμ=1cosx∫00f(μ)dμand the left hand side can also be written as 1cosx∫00f(μ)dμ by substituting μ=sin(x). ■I am not sure if this correct. I may have missed something or if by chance this happens to be correct is there a better proof perhaps?Thanks.

i cannot understand what you want to say, but i have a proof see if you can make me understand that

The second term will be equal to f'(sin x) * cos x using the chain rule. The first term is f'(sin x) . As an example let f (x) be x^2. f'(x) = 2x. Hence, LHS = 2sinx RHS = cos x * 2 sinx

∫baf(x)dx=∫baf(a+b−x)dxIn this case,∫π/20f(sinx)dx=∫π/20f(sin(π/2−x))dx=∫π/20f(cosx)dxThe "identity" is proved easily like so. Let u=a+b−x, then du=−dx. Hence∫baf(x)dx=−∫abf(a+b−u)du=∫baf(a+b−u)du=∫baf(a+b−x)dx