Blog: `int x sin(x^2) dx` Evaluate the indefinite integral.

Friday, April 9, 2010

$\displaystyle\int{x}{\sin{{\left({{x}}^{{2}}\right)}}}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

You need to evaluate the indefinite integral by performing the substitution $\displaystyle{{x}}^{{2}}={t}$ , such that:

$\displaystyle{{x}}^{{2}}={t}\Rightarrow{2}{x}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{x}{\left.{d}{x}\right.}=\frac{{{\left.{d}{t}\right.}}}{{2}}$

$\displaystyle\int{x}\cdot{\sin{{\left({{x}}^{{2}}\right)}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}\int{\sin{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{\left(\frac{{1}}{{2}}\right)}\int{\sin{{t}}}{\left.{d}{t}\right.}={\left(\frac{{1}}{{2}}\right)}{\left(-{\cos{{t}}}\right)}+{c}$

Replacing back $\displaystyle{{x}}^{{2}}$ for t yields:

$\displaystyle\int{x}\cdot{\sin{{\left({{x}}^{{2}}\right)}}}{\left.{d}{x}\right.}=-{\left(\frac{{1}}{{2}}\right)}{\left({{\cos{{x}}}}^{{2}}\right)}+{c}$

Hence, evaluating the indefinite integral yields $\displaystyle\int{x}\cdot{\sin{{\left({{x}}^{{2}}\right)}}}{\left.{d}{x}\right.}=-{\left(\frac{{1}}{{2}}\right)}{\left({{\cos{{x}}}}^{{2}}\right)}+{c}$

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Friday, April 9, 2010

$\displaystyle\int{x}{\sin{{\left({{x}}^{{2}}\right)}}}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

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