maths problem

magpiemagpie46

hi all, given g(x) = f(sin(x)) and g'(x) = -3 cos^2 (x)sin(x)
find f
any help greatly appreciated

Reply 1

mqb2766

Integrate g' and think about how it transforms sin(x)

Reply 2

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

(edited 4 years ago)

Reply 3

magpiemagpie46

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

Reply 4

mqb2766

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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