**Moderator's edit**:

Find a continuous function \(f(x) \) not everywhere zero such that \( \displaystyle (f(x))^2 = \int_0^x \dfrac{f(t) \cdot \sin t}{2 + \cos t} \, dt \).

**Moderator's edit**:

Find a continuous function \(f(x) \) not everywhere zero such that \( \displaystyle (f(x))^2 = \int_0^x \dfrac{f(t) \cdot \sin t}{2 + \cos t} \, dt \).

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## Comments

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TopNewestWhat's the problem?? Just differentiate it to get: \[2 f(x) f'(x)=\dfrac{f(x) \sin x}{2+\cos x}\] Now for f(x)\(\neq\)0, we can write : \[2f'(x)=\dfrac{\sin x}{2+\cos x}\] Now integrate both sides to get f(x) which is very easy... – Rishabh Cool · 10 months ago

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– Akhilesh Prasad · 10 months ago

Thanks a lot for clarifying that out had been banging my head like crazy.Log in to reply

– Akhilesh Prasad · 10 months ago

So it was \(f^{ 2 }\left( x \right) ={ (f\left( x \right) ) }^{ 2 }\) not \(f^{ 2 }\left( x \right) ={ f^{ '' }\left( x \right) }\)Log in to reply

– Rishabh Cool · 10 months ago

Yeah.... Thanks to moderator's edit.... :-}Log in to reply

@Aareyan Manzoor – Akhilesh Prasad · 10 months ago

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@parv mor – Akhilesh Prasad · 10 months ago

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