# Explain why neither Substitution nor Integration by Parts would work to evaluate integral of \cos{{\left({x}^{2}\right)}}{\left.{d}{x}\right.}

Explain why neither Substitution nor Integration by Parts would work to evaluate integral of $$\displaystyle \cos{{\left({x}^{2}\right)}}{\left.{d}{x}\right.}$$

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Nichole Watt

Possible derivation: $$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left( \cos{{\left({x}^{2}\right)}}\right)}$$

Using the chain rule,

$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left( \cos{{\left({x}^{2}\right)}}\right)}=\frac{{{d} \cos{{\left({u}\right)}}}}{{{d}{u}}}\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$$, where $$u = x^2$$ and $$\displaystyle\frac{d}{{{d}{u}}}{\left( \cos{{\left({u}\right)}}\right)}=- \sin{{\left({u}\right)}}:$$

$$\displaystyle=-{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)} \sin{{\left({x}^{2}\right)}}$$

Use the power rule,

$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{n}\right)}={n}{x}^{{{n}-{1}}}$$, where $$n = 2$$. $$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}={2}{x}:$$

$$\displaystyle=- \sin{{\left({x}^{2}\right)}}{2}{x}$$