New Notes Blog: `int xsin^2x dx` Find the indefinite integral

Saturday, 31 October 2015

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

To solve the indefinite integral, we follow $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of f(x)

$\displaystyle{C}$ as the constant of integration.

For the given integral problem: int x sin^2(x) dx, we may apply integration by parts: $\displaystyle\int{u}\cdot{d}{v}={u}{v}-\int{v}\cdot{d}{u}$ .

We may let:

$\displaystyle{u}={x}$ then $\displaystyle{d}{u}={1}{\left.{d}{x}\right.}$ or $\displaystyle{\left.{d}{x}\right.}$

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

To solve the indefinite integral, we follow $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of f(x)

$\displaystyle{C}$ as the constant of integration.

For the given integral problem: int x sin^2(x) dx, we may apply integration by parts: $\displaystyle\int{u}\cdot{d}{v}={u}{v}-\int{v}\cdot{d}{u}$ .

We may let:

$\displaystyle{u}={x}$ then $\displaystyle{d}{u}={1}{\left.{d}{x}\right.}$ or $\displaystyle{\left.{d}{x}\right.}$

$\displaystyle{d}{v}={{\sin}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ then $\displaystyle{v}=\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}$

Note: From the table of integrals, we have $\displaystyle\int{{\sin}}^{{2}}{\left({a}{x}\right)}{\left.{d}{x}\right.}=\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{a}{x}\right)}}}}{{{4}{a}}}$ . We apply this on $\displaystyle{v}=\int{d}{v}=\int{{\sin}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ where $\displaystyle{a}={1}$ .

Applying the formula for integration by parts, we have:

$\displaystyle\int{x}{{\sin}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}={x}\cdot{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}-\int{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{4}}-\int{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}{\left.{d}{x}\right.}$

For the integral: $\displaystyle\int{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}{\left.{d}{x}\right.}$ , we may apply the basic integration property: : $\displaystyle\int{\left({u}-{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}-\int{\left({v}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle\int{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}{\left.{d}{x}\right.}=\int{\left(\frac{{x}}{{2}}\right)}{\left.{d}{x}\right.}-\int\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}{\)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}\int{x}{\left.{d}{x}\right.}-\frac{{1}}{{4}}\int{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}$ .

Apply the Power rule for integration:

$\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{c}$

$\displaystyle\frac{{1}}{{2}}\int{x}{\left.{d}{x}\right.}=\frac{{1}}{{2}}\cdot\frac{{{x}}^{{{1}+{1}}}}{{{1}+{1}}}$

$\displaystyle=\frac{{1}}{{2}}\cdot\frac{{{x}}^{{2}}}{{2}}$

$\displaystyle=\frac{{{x}}^{{2}}}{{4}}$

Apply the basic integration formula for sine function: $\displaystyle\int{\sin{{\left({u}\right)}}}{d}{u}=-{\cos{{\left({u}\right)}}}+{C}$ .

Let: $\displaystyle{u}={2}{x}$ then $\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{2}}={\left.{d}{x}\right.}$ .

$\displaystyle\frac{{1}}{{4}}\int{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}=\frac{{1}}{{4}}\int{\sin{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{2}}$

$\displaystyle=\frac{{1}}{{4}}\cdot\frac{{1}}{{2}}\int{\sin{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\frac{{1}}{{8}}{\left(-{\cos{{\left({u}\right)}}}\right)}$

$\displaystyle=-\frac{{\cos{{\left({u}\right)}}}}{{8}}$

Plug-in $\displaystyle{u}={2}{x}$ on $\displaystyle-\frac{{\cos{{\left({u}\right)}}}}{{8}}$ , we get: $\displaystyle\frac{{1}}{{4}}\int{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}=-\frac{{\cos{{\left({2}{x}\right)}}}}{{8}}$ .

Combining the results, we get:

$\displaystyle\int{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}{\left.{d}{x}\right.}=\frac{{{x}}^{{2}}}{{4}}-{\left(-\frac{{\cos{{\left({2}{x}\right)}}}}{{8}}\right)}+{C}$

$\displaystyle=\frac{{{x}}^{{2}}}{{4}}+\frac{{\cos{{\left({2}{x}\right)}}}}{{8}}+{C}$

Then, the complete indefinite integral will be:

$\displaystyle\int{x}{{\sin}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}=\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{4}}-\int{\left(\frac{{x}}{{2}}-\frac{{\sin{{\left({2}{x}\right)}}}}{{4}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{4}}-{\left(\frac{{{x}}^{{2}}}{{4}}+\frac{{\cos{{\left({2}{x}\right)}}}}{{8}}\right)}+{C}$

$\displaystyle=\frac{{{x}}^{{2}}}{{2}}-\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{4}}-\frac{{{x}}^{{2}}}{{4}}-\frac{{\cos{{\left({2}{x}\right)}}}}{{8}}+{C}$

$\displaystyle=\frac{{{{x}}^{{2}}}}{{4}}-\frac{{{x}{\sin{{\left({2}{x}\right)}}}}}{{4}}-\frac{{\cos{{\left({2}{x}\right)}}}}{{8}}+{C}$

New Notes Blog

Saturday, 31 October 2015

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

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Saturday, 31 October 2015

Find the indefinite integral

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In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

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

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?