Blog: `int (x - 1) sin(pi x ) dx` Evaluate the integral

Wednesday, December 4, 2013

$\displaystyle\int{\left({x}-{1}\right)}{\sin{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

You need to solve the integral $\displaystyle\int{\left({x}-{1}\right)}{\sin{{\left(\pi\cdot{x}\right)}}}{\left.{d}{x}\right.}=\int{x}\cdot{\sin{{\left(\pi\cdot{x}\right)}}}{\left.{d}{x}\right.}-\int{\sin{{\left(\pi\cdot{x}\right)}}}{\left.{d}{x}\right.}$

You need to use substitution $\displaystyle\pi\cdot{x}={t}\Rightarrow\pi\cdot{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{\left.{d}{x}\right.}=\frac{{{\left.{d}{t}\right.}}}{{\pi}}$

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

You need to use the integration by parts for $\displaystyle\int{t}\cdot{\sin{{t}}}$ such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={t}\Rightarrow{d}{u}={\left.{d}{t}\right.}$

$\displaystyle{d}{v}={\sin{{t}}}\Rightarrow{v}=-{\cos{{t}}}$

...

You need to use substitution $\displaystyle\pi\cdot{x}={t}\Rightarrow\pi\cdot{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{\left.{d}{x}\right.}=\frac{{{\left.{d}{t}\right.}}}{{\pi}}$

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

You need to use the integration by parts for $\displaystyle\int{t}\cdot{\sin{{t}}}$ such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={t}\Rightarrow{d}{u}={\left.{d}{t}\right.}$

$\displaystyle{d}{v}={\sin{{t}}}\Rightarrow{v}=-{\cos{{t}}}$

$\displaystyle\int{t}\cdot{\sin{{t}}}=-{t}\cdot{\cos{{t}}}+\int{\cos{{t}}}{\left.{d}{t}\right.}$

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

Replacing back the variable yields:

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

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

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

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Wednesday, December 4, 2013

$\displaystyle\int{\left({x}-{1}\right)}{\sin{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Wednesday, December 4, 2013

Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\int{\left({x}-{1}\right)}{\sin{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}$ Evaluate the integral

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?