let $$f(x)=\int_{1}^{x}{\frac{\cos y\sin x}{y^2+y+1}dy}$$ then find all values of $x$ for which $f'(x)=0$?

I tried directly differentiating but I got stuck on $$f'(x)=\cos x\left(\int_1^x{\frac{\cos y}{y^2+y+1}dy}+\frac{\sin x}{x^2+x+1}\right)$$

  • $\begingroup$ The last integral is totally wrong $\endgroup$ Sep 22 at 14:23
  • $\begingroup$ @MathAttack I did $f(x)=\sin x\int_1^x{\frac{cosy}{y^2+y+1}dy}$ then differentiated by product rule what did I do wrong? $\endgroup$
    – Scilent144
    Sep 22 at 14:29
  • $\begingroup$ @Gonçalo I edited the question $\endgroup$
    – Scilent144
    Sep 22 at 14:38

1 Answer 1


Given: $$f(x)=\int_{1}^{x} \frac{\cos y \sin x}{y^2+y+1} dy $$

Take the derivative (Using the product rule for the RHS): $$\implies \frac{d}{dx}f(x)=\frac{d}{dx}[\sin(x)\int_{1}^{x} \frac{\cos y}{y^2+y+1} dy]$$ $$\implies f'(x)=\cos(x)\int_{1}^{x} \frac{\cos y}{y^2+y+1}dy +\frac{\cos x}{x^2+x+1}\sin x$$ Now we want to find all values of x such that $f'(x)=0$: $$\cos x \int_{1}^{x}\frac{\cos y}{y^2+y+1} dy +\frac{\sin x \cos x}{x^2+x+1}=0$$ Well, notice that the integral will not be zero simultaneously with the other expression (As if $x=1$ (to make the integral zero), then the other expression is not zero; and if there exists an $x$ such that the integral is zero, the other expression won't be zero. Hence, it seems the only way to make this equation to be zero is to set $cos(x)=0$ and so $x=(2n+1)\frac{\pi}{2}$ is the solution for all integers $n$.

  • 2
    $\begingroup$ Though, I am not sure 100% of my answer. So make sure either from the final solutions in the book or problem set you have, or if someone can confirm, please do so. $\endgroup$
    – Nero
    Sep 22 at 14:45
  • 1
    $\begingroup$ yeah your answer is right. the question was multiple choice so the question is kinda wrong question is actually asking for all x for which f'(x) is definitely 0 and not for all zeroes so sorry about that $\endgroup$
    – Scilent144
    Sep 22 at 14:51

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