`int_0^a (dx)/(a^2 + x^2)^(3/2) , a>0` Evaluate the integral

Friday, September 6, 2013

$\displaystyle{\int_{{0}}^{{a}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({{a}}^{{2}}+{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}},{a}>{0}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{a}}}\frac{{\left.{d}{x}}}{{{\left({{a}}^{{2}}+{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}$

Let's first evaluate the indefinite integral by integral substitution,

Let $\displaystyle{x}={a}{\tan{{\left({u}\right)}}}$

$\displaystyle{\left.{d}{x}\right.}={a}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

$\displaystyle\int\frac{{\left.{d}{x}}}{{{\left({{a}}^{{2}}+{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}=\int\frac{{{a}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}}}{{{\left({{a}}^{{2}}+{{a}}^{{2}}{{\tan}}^{{2}}{\left({u}\right)}\right)}}^{{\frac{{3}}{{2}}}}}$

$\displaystyle=\int\frac{{{a}{{\sec}}^{{2}}{\left({u}\right)}}}{{{\left({{a}}^{{2}}{\left({1}+{{\tan}}^{{2}}{\left({u}\right)}\right)}\right)}}^{{\frac{{3}}{{2}}}}}{d}{u}$

$\displaystyle=\int\frac{{{a}{{\sec}}^{{2}}{\left({u}\right)}}}{{{{\left({{a}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}{{\left({1}+{{\tan}}^{{2}}{\left({u}\right)}\right)}}^{{\frac{{3}}{{2}}}}}}{d}{u}$

Use the identity: $\displaystyle{1}+{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int\frac{{{a}{{\sec}}^{{2}}{\left({u}\right)}}}{{{\left({{a}}^{{3}}\right)}{{\left({{\sec}}^{{2}}{\left({u}\right)}\right)}}^{{\frac{{3}}{{2}}}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{{a}}^{{2}}}\int\frac{{{{\sec}}^{{2}}{\left({u}\right)}}}{{{{\sec}}^{{3}}{\left({u}\right)}}}{d}{u}$

$\displaystyle=\frac{{1}}{{{a}}^{{2}}}\int\frac{{1}}{{\sec{{\left({u}\right)}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{{a}}^{{2}}}\int{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\frac{{1}}{{{a}}^{{2}}}{\left({\sin{{\left({u}\right)}}}\right)}$

We have used $\displaystyle{x}={a}{\tan{{\left({u}\right)}}}$

$\displaystyle{\tan{{\left({u}\right)}}}=\frac{{x}}{{a}}$

Now let's find sin(u) for triangle with angle theta, opposite side as x and adjacent side as a and hypotenuse as h,

$\displaystyle{{h}}^{{2}}={{x}}^{{2}}+{{a}}^{{2}}$

$\displaystyle{h}=\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}$

So, $\displaystyle{\sin{{\left({u}\right)}}}=\frac{{x}}{\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}$

$\displaystyle=\frac{{1}}{{{a}}^{{2}}}{\left(\frac{{x}}{\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}\right)}$

Add a constant C to the solution.

$\displaystyle=\frac{{1}}{{{a}}^{{2}}}{\left(\frac{{x}}{\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}\right)}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{a}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({{a}}^{{2}}+{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}={{\left[\frac{{1}}{{{a}}^{{2}}}{\left(\frac{{x}}{\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}\right)}\right]}_{{0}}^{{a}}}$

$\displaystyle={\left[\frac{{1}}{{{a}}^{{2}}}{\left(\frac{{a}}{\sqrt{{{{a}}^{{2}}+{{a}}^{{2}}}}}\right)}\right]}-{\left[\frac{{1}}{{{a}}^{{2}}}{\left(\frac{{0}}{\sqrt{{{{0}}^{{2}}+{{a}}^{{2}}}}}\right)}\right]}$

$\displaystyle={\left[\frac{{1}}{{{a}\sqrt{{{2}{{a}}^{{2}}}}}}\right]}-{\left[{0}\right]}$

$\displaystyle={\left(\frac{{1}}{{{{a}}^{{2}}\sqrt{{{2}}}}}\right)}$

$\displaystyle=\frac{{1}}{{\sqrt{{{2}}}{{a}}^{{2}}}}$

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Friday, September 6, 2013

$\displaystyle{\int_{{0}}^{{a}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({{a}}^{{2}}+{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}},{a}>{0}$ Evaluate the integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Friday, September 6, 2013

Evaluate the integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{0}}^{{a}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({{a}}^{{2}}+{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}},{a}>{0}$ Evaluate the integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?