A function f satisfying f'( sin x ) = cos^2 x | vedclass

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(A) Given $f'(\sin x) = \cos^2 x$. Let $t = \sin x$,then $\cos^2 x = 1 - \sin^2 x = 1 - t^2$. So,$f'(t) = 1 - t^2$. Integrating both sides with respec

Vedclass · Accepted Answer

$f(x) = x - \frac{x^3}{3} + \frac{1}{3}$

Vedclass · Answer

(A) Given $f'(\sin x) = \cos^2 x$. Let $t = \sin x$,then $\cos^2 x = 1 - \sin^2 x = 1 - t^2$. So,$f'(t) = 1 - t^2$. Integrating both sides with respect to $t$,we get: $f(t) = \int (1 - t^2) dt = t - \frac{t^3}{3} + C$. Given $f(1) = 1$,substitute $t = 1$: $1 = 1 - \frac{1^3}{3} + C \Rightarrow 1 = 1 - \frac{1}{3} + C \Rightarrow C = \frac{1}{3}$. Thus,$f(t) = t - \frac{t^3}{3} + \frac{1}{3}$. Replacing $t$ with $x$,we get $f(x) = x - \frac{x^3}{3} + \frac{1}{3}$.

$A$ function $f$ satisfying $f'( \sin x ) = \cos^2 x$ for all $x$ and $f(1) = 1$ is :

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