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Find $f(x)$ that satisfies the first-order differential equation: $$f'(\sin^2x)=\cos^2x+\tan^2x\tag{for $0

2 Answers 2

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Hint: We can easily observe that $$f'(\sin^2 x) = 1- \sin^2 x + \frac {\sin^2 x}{1-\sin^2 x} $$ We can thus generalise to $$f'(y) = 1- y + \frac {y}{1-y} \tag {1} $$ which can be solved for $f (x) $ by simply integrating $(1)$. Hope it helps.

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Hint: Show that the given differential equation can be represented as $$f'(x)=1-x+\frac{x}{1-x}$$ and then solve this by simply integrating.

Hope this helps you.

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    If $x =\sin^2 y$, then sir your expression takes the fourth powers of $\sin $. – 2017-02-01
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    @Chinny84 Thanks for mentioning. I was on the verge of correcting it though. – 2017-02-01