# (Integration) Question on U-Substitution Mistake

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.

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

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