prove that
f′(sinx)=(f(sinx))′
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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.
Utsavsterbon:
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
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
thanks Bro
you understood?
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∫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
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