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

Friday, 15 November 2013

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

Given to solve,

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

using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows

for $\displaystyle\sqrt{{{a}+{b}{{x}}^{{2}}}}$ we can take $\displaystyle{x}=\sqrt{{\frac{{a}}{{b}}}}{\tan{{\left({u}\right)}}}$

so ,For

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

the $\displaystyle{x}=\sqrt{{\frac{{4}}{{1}}}}{\tan{{\left({u}\right)}}}={2}{\tan{{\left({u}\right)}}}$

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

so,

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

= $\displaystyle\int\sqrt{{{4}+{{\left({2}{\tan{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{{\sec}}^{{{2}}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{4}+{4}{{\left({\tan{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{{\sec}}^{{{2}}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{4}{\left({1}+{{\left({\tan{{\left({u}\right)}}}\right)}}^{{2}}\right)}}}{\left({2}{{\sec}}^{{{2}}}{\left({u}\right)}{d}{u}\right)}$

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

...

Given to solve,

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

using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows

for $\displaystyle\sqrt{{{a}+{b}{{x}}^{{2}}}}$ we can take $\displaystyle{x}=\sqrt{{\frac{{a}}{{b}}}}{\tan{{\left({u}\right)}}}$

so ,For

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

the $\displaystyle{x}=\sqrt{{\frac{{4}}{{1}}}}{\tan{{\left({u}\right)}}}={2}{\tan{{\left({u}\right)}}}$

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

so,

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

= $\displaystyle\int\sqrt{{{4}+{{\left({2}{\tan{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{{\sec}}^{{{2}}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{4}+{4}{{\left({\tan{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({2}{{\sec}}^{{{2}}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle\int\sqrt{{{4}{\left({1}+{{\left({\tan{{\left({u}\right)}}}\right)}}^{{2}}\right)}}}{\left({2}{{\sec}}^{{{2}}}{\left({u}\right)}{d}{u}\right)}$

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

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

= $\displaystyle\int{4}{{\sec}}^{{{3}}}{\left({u}\right)}{d}{u}$

$\displaystyle={4}\int{{\sec}}^{{{3}}}{\left({u}\right)}{d}{u}$

by applying the Integral Reduction

$\displaystyle\int{{\sec}}^{{{n}}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{{\sec}}^{{{n}-{1}}}{\left({x}\right)}{\sin{{\left({x}\right)}}}}}{{{n}-{1}}}+{\left(\frac{{{n}-{2}}}{{{n}-{1}}}\right)}\int{{\sec}}^{{{n}-{2}}}{\left({x}\right)}{\left.{d}{x}\right.}$

so ,

$\displaystyle{4}\int{{\sec}}^{{{3}}}{\left({u}\right)}{d}{u}$

= $\displaystyle{4}{\left[\frac{{{{\sec}}^{{{3}-{1}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{3}-{1}}}+{\left(\frac{{{3}-{2}}}{{{3}-{1}}}\right)}\int{{\sec}}^{{{3}-{2}}}{\left({u}\right)}{d}{u}\right]}$

= $\displaystyle{4}{\left[\frac{{{{\sec}}^{{{2}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{2}}}+{\left(\frac{{{1}}}{{{2}}}\right)}\int{\sec{{\left({u}\right)}}}{d}{u}\right]}$

= $\displaystyle{4}{\left[\frac{{{{\sec}}^{{{2}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{\left({\ln{{\left({\tan{{\left({u}\right)}}}+{\sec{{\left({u}\right)}}}\right)}}}\right)}\right]}$

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

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

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

so,

$\displaystyle{4}{\left[\frac{{{{\sec}}^{{{2}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{\left({\ln{{\left({\tan{{\left({u}\right)}}}+{\sec{{\left({u}\right)}}}\right)}}}\right)}\right]}$

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

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

New Notes Blog

Friday, 15 November 2013

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

Friday, 15 November 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?

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