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For context, I was playing with this integral recreationally: $$\int{20\sin(\frac{x^2}{35})}dx$$ I decided to try u-substitution, and got the following: $$u=\frac{x^2}{35}\space,\space du=\frac{2x}{35}dx\space,\space x=\pm\sqrt{35u}$$ $$\frac{35}{2x}du=dx\space,\space \pm\frac{35}{2(\sqrt{35u})}du=dx$$ Substituting $u$ for $\frac{x^2}{35}$ and $\pm\frac{35}{2(\sqrt{35u})}du$ for $dx$, I rewrite the integral as: $$\pm\frac{20*35}{2\sqrt{35}}\int{\sin(u)\frac{1}{\sqrt{u}}}du$$ Just to check, I evaluated both from $0$ to $\pi$ absolutely in Mathematica, and obtained: $$|\int^{\pi}_{0}{20\sin(\frac{x^2}{35})}dx|\approx5.8725 \space , \space |\pm\frac{20*35}{2\sqrt{35}}\int^{\pi}_{0}{\sin(u)\frac{1}{\sqrt{u}}}du|\approx105.88$$ Getting different numerical approximations, I assume I did something wrong. So my question is what was my mistake?

Please keep in mind I don't need to symbolically solve this integral, I just wish to better understand symbol pushing.

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    $\begingroup$ @Atmos bijectivity is not at all a requirement for $u$-substitution $\endgroup$ Jan 20 at 20:12

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You simply forgot to change the bounds.... $$ \frac{20\times 35}{2\sqrt{35}}\int_{0}^{\frac{\pi^2}{35}}\frac{\sin(u)}{\sqrt{u}}du \approx 5.872 $$

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  • $\begingroup$ Oh boy do I feel silly. Thanks for answering! $\endgroup$
    – Chloe
    Jan 20 at 20:15
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    $\begingroup$ @Chloe it's all good! I make many mistakes like this all the time too! $\endgroup$ Jan 20 at 20:16

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