New Notes Blog: `int x^2/sqrt(36-x^2) dx` Find the indefinite integral

Thursday, 10 April 2014

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Given

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

This can be solved by using the Trigonometric substitutions (Trig substitutions)

when the integral contains $\displaystyle\sqrt{{{a}-{b}{{x}}^{{2}}}}$ then we have to take

$\displaystyle{x}=\sqrt{{\frac{{a}}{{b}}}}{\sin{{\left({t}\right)}}}$ in order to solve the integral easily

so here , For

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

$\displaystyle{x}$ is given as

$\displaystyle{x}=\sqrt{{\frac{{36}}{{1}}}}{\sin{{\left({t}\right)}}}={6}{\sin{{\left({t}\right)}}}$

=> $\displaystyle{\left.{d}{x}\right.}={6}{\cos{{\left({t}\right)}}}{\left.{d}{t}\right.}$

so ,

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{{\left({6}{\sin{{\left({t}\right)}}}\right)}}^{{2}}}{\sqrt{{{36}-{{\left({6}{\sin{{\left({t}\right)}}}\right)}}^{{2}}}}}{\left({6}{\cos{{\left({t}\right)}}}{\left.{d}{t}\right.}\right)}$

= $\displaystyle\int{36}\frac{{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}}{\sqrt{{{36}-{{\left({6}{\sin{{\left({t}\right)}}}\right)}}^{{2}}}}}{\left({6}{\cos{{\left({t}\right)}}}{\left.{d}{t}\right.}\right)}$

= $\displaystyle\ldots$

Given

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

This can be solved by using the Trigonometric substitutions (Trig substitutions)

when the integral contains $\displaystyle\sqrt{{{a}-{b}{{x}}^{{2}}}}$ then we have to take

$\displaystyle{x}=\sqrt{{\frac{{a}}{{b}}}}{\sin{{\left({t}\right)}}}$ in order to solve the integral easily

so here , For

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

$\displaystyle{x}$ is given as

$\displaystyle{x}=\sqrt{{\frac{{36}}{{1}}}}{\sin{{\left({t}\right)}}}={6}{\sin{{\left({t}\right)}}}$

=> $\displaystyle{\left.{d}{x}\right.}={6}{\cos{{\left({t}\right)}}}{\left.{d}{t}\right.}$

so ,

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\frac{{{\left({36}\right)}\cdot{\left({6}\right)}{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}\cdot{\cos{{\left({t}\right)}}}}}{\sqrt{{{36}-{{\left({6}{\sin{{\left({t}\right)}}}\right)}}^{{2}}}}}{\left.{d}{t}\right.}$

= $\displaystyle\int\frac{{{216}{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}\cdot{\cos{{\left({t}\right)}}}}}{\sqrt{{{36}-{36}{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}}}}{\left.{d}{t}\right.}$

= $\displaystyle\int\frac{{{216}{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}\cdot{\cos{{\left({t}\right)}}}}}{\sqrt{{{36}{\left({1}-{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}\right)}}}}{\left.{d}{t}\right.}$

= $\displaystyle\int\frac{{{216}{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}\cdot{\cos{{\left({t}\right)}}}}}{\sqrt{{{36}{{\left({\cos{{\left({t}\right)}}}\right)}}^{{2}}}}}{\left.{d}{t}\right.}$

= $\displaystyle\int\frac{{{216}{{\left({\sin{{\left({t}\right)}}}\right)}}^{{2}}\cdot{\cos{{\left({t}\right)}}}}}{{{6}{\left({\cos{{\left({t}\right)}}}\right)}}}{\left.{d}{t}\right.}$

= $\displaystyle\int{\left(\frac{{216}}{{6}}\right)}{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

= $\displaystyle\int{36}{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

= $\displaystyle{36}\int{{\sin}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

= $\displaystyle{36}\int\frac{{{1}-{\cos{{\left({2}{t}\right)}}}}}{{2}}{\left.{d}{t}\right.}$

= $\displaystyle{\left(\frac{{36}}{{2}}\right)}\int{\left({1}-{\cos{{\left({2}{t}\right)}}}\right)}{\left.{d}{t}\right.}$

= $\displaystyle{18}{\left[\int{1}{\left.{d}{t}\right.}-\int{\cos{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}\right]}+{c}$

= $\displaystyle{18}{\left[{t}-{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{t}\right)}}}\right]}+{c}$

but we know that

$\displaystyle{x}={6}{\sin{{\left({t}\right)}}}$

=> $\displaystyle\frac{{x}}{{6}}={\sin{{\left({t}\right)}}}$

=> $\displaystyle{t}={{\sin}}^{{-{1}}}{\left(\frac{{x}}{{6}}\right)}{\quad\text{or}\quad}{\arcsin{{\left(\frac{{x}}{{6}}\right)}}}$

so,

$\displaystyle{18}{\left[{t}-{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{t}\right)}}}\right]}+{c}$

= $\displaystyle{18}{\left[{\left({\arcsin{{\left(\frac{{x}}{{6}}\right)}}}\right)}-{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{\left({\arcsin{{\left(\frac{{x}}{{6}}\right)}}}\right)}\right)}}}\right]}+{c}$

so,

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

= $\displaystyle{18}{\arcsin{{\left(\frac{{x}}{{6}}\right)}}}-{9}{\sin{{\left({2}{\left({\arcsin{{\left(\frac{{x}}{{6}}\right)}}}\right)}\right)}}}+{c}$

New Notes Blog

Thursday, 10 April 2014

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ 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?

Thursday, 10 April 2014

Find the indefinite integral

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Post a Comment

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?

$\displaystyle\int\frac{{{x}}^{{2}}}{\sqrt{{{36}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ 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?