New Notes Blog: `int sqrt(16-4x^2)dx` Find the indefinite integral

Monday, 16 September 2013

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Given ,

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

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

For $\displaystyle\sqrt{{{a}-{b}{{x}}^{{2}}}}$ we have to take $\displaystyle{x}=$ $\displaystyle\sqrt{{\frac{{a}}{{b}}}}{\sin{{\left({u}\right)}}}$

so here , For

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}-----{\left({1}\right)}$

$\displaystyle{x}$ can be given as

$\displaystyle{x}=\sqrt{{\frac{{16}}{{4}}}}{\sin{{\left({u}\right)}}}=\sqrt{{{4}}}{\sin{{\left({u}\right)}}}={2}{\sin{{\left({u}\right)}}}$

so, $\displaystyle{x}={2}{\sin{{\left({u}\right)}}}$ => $\displaystyle{\left.{d}{x}\right.}={2}{\cos{{\left({u}\right)}}}{d}{u}$

Now substituting $\displaystyle{x}$ in (1) we get,

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\sqrt{{{16}-{4}{{\left({2}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{16}-{4}\cdot{4}{{\left({\sin{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

...

Given ,

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

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

For $\displaystyle\sqrt{{{a}-{b}{{x}}^{{2}}}}$ we have to take $\displaystyle{x}=$ $\displaystyle\sqrt{{\frac{{a}}{{b}}}}{\sin{{\left({u}\right)}}}$

so here , For

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}-----{\left({1}\right)}$

$\displaystyle{x}$ can be given as

$\displaystyle{x}=\sqrt{{\frac{{16}}{{4}}}}{\sin{{\left({u}\right)}}}=\sqrt{{{4}}}{\sin{{\left({u}\right)}}}={2}{\sin{{\left({u}\right)}}}$

so, $\displaystyle{x}={2}{\sin{{\left({u}\right)}}}$ => $\displaystyle{\left.{d}{x}\right.}={2}{\cos{{\left({u}\right)}}}{d}{u}$

Now substituting $\displaystyle{x}$ in (1) we get,

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int\sqrt{{{16}-{4}{{\left({2}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{16}-{4}\cdot{4}{{\left({\sin{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{16}-{16}{{\left({\sin{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{16}{\left({1}-{{\left({\sin{{\left({u}\right)}}}\right)}}^{{2}}\right)}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{16}{{\left({\cos{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int{\left({4}{\cos{{\left({u}\right)}}}\right)}{\left({2}{\cos{{\left({u}\right)}}}{d}{u}\right)}$

= $\displaystyle\int{8}{{\cos}}^{{2}}{\left({u}\right)}{d}{u}$

= $\displaystyle{8}\int{{\cos}}^{{2}}{\left({u}\right)}{d}{u}$

= $\displaystyle{8}\int\frac{{{1}+{\cos{{\left({2}{u}\right)}}}}}{{2}}{d}{u}$

= $\displaystyle{\left(\frac{{8}}{{2}}\right)}\int{\left({1}+{\cos{{\left({2}{u}\right)}}}\right)}{d}{u}$

= $\displaystyle{4}\int{\left({1}+{\cos{{\left({2}{u}\right)}}}\right)}{d}{u}$

= $\displaystyle{4}{\left[\int{1}{d}{u}+\int{\cos{{\left({2}{u}\right)}}}{d}{u}\right]}$

= $\displaystyle{4}{\left[{u}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{u}\right)}}}\right)}\right]}+{c}$

but $\displaystyle{x}={2}{\sin{{\left({u}\right)}}}$

=> $\displaystyle{\left(\frac{{x}}{{2}}\right)}={\sin{{\left({u}\right)}}}$

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

so,

$\displaystyle{4}{\left[{u}+{\left(\frac{{1}}{{2}}\right)}{\left({\sin{{\left({2}{u}\right)}}}\right)}\right]}+{c}$

= $\displaystyle{4}{\left[{{\sin}}^{{-{1}}}{\left(\frac{{x}}{{2}}\right)}+\frac{{1}}{{2}}{\sin{{\left({2}{\left({{\sin}}^{{-{1}}}{\left(\frac{{x}}{{2}}\right)}\right)}\right)}}}\right]}+{c}$

so,

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle{4}{{\sin}}^{{-{1}}}{\left(\frac{{x}}{{2}}\right)}+{2}{\sin{{\left({2}{\left({{\sin}}^{{-{1}}}{\left(\frac{{x}}{{2}}\right)}\right)}\right)}}}+{c}$

New Notes Blog

Monday, 16 September 2013

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ 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?

Monday, 16 September 2013

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?

$\displaystyle\int\sqrt{{{16}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ 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?