New Notes Blog: `int sqrt(9+16x^2) dx` Find the indefinite integral

Tuesday, 22 October 2013

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Given to solve,

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

by using the trig substitution , we can solve the integral

for $\displaystyle\sqrt{{{a}+{b}{{x}}^{{2}}}}$ $\displaystyle{\left.{d}{x}\right.}$ the $\displaystyle{x}$ is given as

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

so,

for the integral

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

let $\displaystyle{x}=\sqrt{{\frac{{9}}{{16}}}}{\tan{{\left({u}\right)}}}={\left(\frac{{3}}{{4}}\right)}{\tan{{\left({u}\right)}}}$

=> $\displaystyle{\left.{d}{x}\right.}={\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

so,

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int{\left[\sqrt{{{9}{\left({1}+\frac{{16}}{{9}}{{x}}^{{2}}\right)}}}\right]}{\left({\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle{3}\int{\left[\sqrt{{{1}+{\left(\frac{{16}}{{9}}\right)}{{x}}^{{2}}}}\right]}{\left({\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle{3}\int\sqrt{{{1}+{\left(\frac{{16}}{{9}}\right)}{{\left({\left(\frac{{3}}{{4}}\right)}{\tan{{\left({u}\right)}}}\right)}}^{{2}}}}{\left({\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

...

Given to solve,

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

by using the trig substitution , we can solve the integral

for $\displaystyle\sqrt{{{a}+{b}{{x}}^{{2}}}}$ $\displaystyle{\left.{d}{x}\right.}$ the $\displaystyle{x}$ is given as

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

so,

for the integral

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

let $\displaystyle{x}=\sqrt{{\frac{{9}}{{16}}}}{\tan{{\left({u}\right)}}}={\left(\frac{{3}}{{4}}\right)}{\tan{{\left({u}\right)}}}$

=> $\displaystyle{\left.{d}{x}\right.}={\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

so,

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

= $\displaystyle\int{\left[\sqrt{{{9}{\left({1}+\frac{{16}}{{9}}{{x}}^{{2}}\right)}}}\right]}{\left({\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle{3}\int{\left[\sqrt{{{1}+{\left(\frac{{16}}{{9}}\right)}{{x}}^{{2}}}}\right]}{\left({\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle{3}\int{\left[\sqrt{{{1}+{\left(\frac{{16}}{{9}}\right)}{\left(\frac{{9}}{{16}}\right)}{\left({{\tan}}^{{2}}{u}\right)}}}\right]}{\left({\left(\frac{{3}}{{4}}\right)}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}\int\sqrt{{{1}+{{\tan}}^{{2}}{u}}}{\left({{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}\int\sqrt{{{{\sec}}^{{2}}{u}}}{\left({{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

$\displaystyle={\left(\frac{{9}}{{4}}\right)}\int{\sec{{u}}}{\left({{\sec}}^{{2}}{\left({u}\right)}{d}{u}\right)}$

$\displaystyle={\left(\frac{{9}}{{4}}\right)}\int{\left({{\sec}}^{{3}}{\left({u}\right)}{d}{u}\right)}$

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 ,

$\displaystyle{\left(\frac{{9}}{{4}}\right)}\int{{\sec}}^{{{3}}}{\left({u}\right)}{d}{u}$

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}{\left[\frac{{{{\sec}}^{{{3}-{1}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{3}-{1}}}+{\left(\frac{{{3}-{2}}}{{{3}-{1}}}\right)}\int{{\sec}}^{{{3}-{2}}}{\left({u}\right)}{d}{u}\right]}$

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}{\left[\frac{{{{\sec}}^{{{2}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{2}}}+{\left(\frac{{{1}}}{{{2}}}\right)}\int{\sec{{\left({u}\right)}}}{d}{u}\right]}$

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}{\left[\frac{{{{\sec}}^{{{2}}}{\left({u}\right)}{\sin{{\left({u}\right)}}}}}{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{\left({\ln{{\left({\tan{{\left({u}\right)}}}+{\sec{{\left({u}\right)}}}\right)}}}\right)}\right]}$

but we know

$\displaystyle{x}={\left(\frac{{3}}{{4}}\right)}{\tan{{\left({u}\right)}}}$

= > $\displaystyle{4}\frac{{x}}{{3}}={\tan{{\left({u}\right)}}}$

=> $\displaystyle{u}={\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}$

so,

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}{\left[\frac{{{{\sec}}^{{{2}}}{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}{\sin{{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}}}}}{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{\left({\ln{{\left({\tan{{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}}}+{\sec{{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}}}\right)}}}\right)}\right]}$

= $\displaystyle{\left(\frac{{9}}{{4}}\right)}{\left[\frac{{{{\sec}}^{{{2}}}{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}{\sin{{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}}}}}{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{\left({\ln{{\left({\left({4}\frac{{x}}{{3}}\right)}\right)}}}+{\sec{{\left({\arctan{{\left({4}\frac{{x}}{{3}}\right)}}}\right)}}}\right)}\right]}+{c}$

New Notes Blog

Tuesday, 22 October 2013

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ 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?

Tuesday, 22 October 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?

$\displaystyle\int\sqrt{{{9}+{16}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ 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?