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I'm having trouble with calculatig this specific integral: $$\int_1^\infty \dfrac{\sin^2x}{\sqrt{1 + x^3}}$$ Any piece of advice on how to get this one done?

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    Bound above by an easy-to-integrate function. – 2017-02-03
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    Hint: $\left | \dfrac{\sin^2x}{\sqrt{1 + x^3}} \right |\le \frac{1}{x^{3/2}}$ – 2017-02-03
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    By *calculating* you mean computing a closed form for such integral, or just proving it is convergent? One problem is intractable, the other one is trivial. – 2017-02-03
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    nah, it was about proving it was convergent. I got it now, thanks. – 2017-02-03

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$$\frac{\sin^2x}{\sqrt{1+x^3}}\le\frac1{x^{3/2}}\;\;,\;\;\;\text{and since}\;\;\int_1^\infty x^pdx\;\;\text{converges}\;\;\iff p<-1$$

then....

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    Is it obvious that this inequality holds? – 2017-02-03
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    @Fabian Well, we know $\;|\sin x|\le 1\implies \sin^2x\le 1\;$ , and in the **positive** ray $\;[1,\infty)\;$ , we have that $\;\sqrt{1+x^3}\ge\sqrt{x^3}\;$ , so I'd rather say it is pretty trivial. – 2017-02-03
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    @Fabian Yes, it is. And (+1) for the concise answer. – 2017-02-03
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The denominator is always positive for $x\in[1,\infty)$, so the only potential problem is when $x\rightarrow\infty$.

However, we have $\sin^2(x)\leq1$ and since $\frac{1}{\sqrt{1+x^3}}<\frac{1}{x^{3/2}}$ is integrable on $[1,\infty)$, your integral is convergent.