magpiemagpie46
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hi all, given g(x) = f(sin(x)) and g'(x) = -3 cos^2 (x)sin(x)
find f
any help greatly appreciated
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mqb2766
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Integrate g' and think about how it transforms sin(x)
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max271
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Integrating g^\prime(x), we find that g(x)=\cos^3(x)+C, where C is a constant.
We know that \cos(x)=\pm\sqrt{1-\sin^2(x)}, where the sign depends on the sign of \cos(x).
Therefore g(x)=\pm\big(1-\sin^2(x)\big)^\frac{3}{2}+C, and consequently f(x)=\pm\big(1-x^2\big)^\frac{3}{2}+C (just be careful when determining which sign belong to which values of x).
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magpiemagpie46
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Thank you
(Original post by max271)
Integrating g^\prime(x), we find that g(x)=\cos^3(x)+C, where C is a constant.
We know that \cos(x)=\pm\sqrt{1-\sin^2(x)}, where the sign depends on the sign of \cos(x).
Therefore g(x)=\pm\big(1-\sin^2(x)\big)^\frac{3}{2}+C, and consequently f(x)=\pm\big(1-x^2\big)^\frac{3}{2}+C (just be careful when determining which sign belong to which values of x).
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mqb2766
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The sticky at the top of the forum says to give hintd, not solutions.
(Original post by max271)
Integrating g^\prime(x), we find that g(x)=\cos^3(x)+C, where C is a constant.
We know that \cos(x)=\pm\sqrt{1-\sin^2(x)}, where the sign depends on the sign of \cos(x).
Therefore g(x)=\pm\big(1-\sin^2(x)\big)^\frac{3}{2}+C, and consequently f(x)=\pm\big(1-x^2\big)^\frac{3}{2}+C (just be careful when determining which sign belong to which values of x).
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