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

Albarellak 2021-08-31 Answered

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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Expert Answer

Nichole Watt
Answered 2021-09-01 Author has 9644 answers

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 \(n = 2\). \(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}={2}{x}:\)

Answer:

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

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