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Integration with Square Root
- Thread starter Yuuki
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mente oscura
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- Nov 29, 2013
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Re: integration
It is possible, that there is an easier way of solving the integral. But it could also be determined:
[tex]\displaystyle \int \ (x^2+2x+1) \sqrt{4x^2+8x+5}[/tex]
[tex]\displaystyle \int \ (x+1)(x+1) \sqrt{4(x+1)^2+1}[/tex]
[tex]u=x+1[/tex]
[tex]v'=(x+1) \sqrt{4(x+1)^2+1}[/tex]
[tex]\displaystyle \ (uv)'=u'v+uv'[/tex]
[tex]\displaystyle \int \ (uv)'= \int \ u'v+ \int \ uv'[/tex]
Therefore:
[tex]\displaystyle \int \ uv' = uv- \int \ u'v[/tex]
Try to follow.
Regards.
Hello.
It is possible, that there is an easier way of solving the integral. But it could also be determined:
[tex]\displaystyle \int \ (x^2+2x+1) \sqrt{4x^2+8x+5}[/tex]
[tex]\displaystyle \int \ (x+1)(x+1) \sqrt{4(x+1)^2+1}[/tex]
[tex]u=x+1[/tex]
[tex]v'=(x+1) \sqrt{4(x+1)^2+1}[/tex]
[tex]\displaystyle \ (uv)'=u'v+uv'[/tex]
[tex]\displaystyle \int \ (uv)'= \int \ u'v+ \int \ uv'[/tex]
Therefore:
[tex]\displaystyle \int \ uv' = uv- \int \ u'v[/tex]
Try to follow.
Regards.
- Thread starter
- #3
Re: integration
[tex]\displaystyle \begin{align*} x + 1 = \frac{1}{2}\tan{\left( \theta \right) } \implies dx = \frac{1}{2} \sec^2{ \left( \theta \right) } \, d\theta \end{align*}[/tex] and the integral becomes
[tex]\displaystyle \begin{align*} \int{ \left( x + 1 \right) ^2 \sqrt{4 \left( x + 1 \right) ^2 + 1 } \, dx} &= \int{ \left[ \frac{1}{2}\tan{ \left( \theta \right) } \right] ^2 \sqrt{ 4 \left[ \frac{1}{2} \tan{ \left( \theta \right) } \right] ^2 + 1 } \, \frac{1}{2} \sec^2{ \left( \theta \right) } \, d\theta } \\ &= \frac{1}{2} \int{ \frac{1}{4}\tan^2{ \left( \theta \right) } \sec^2{ \left( \theta \right) } \, \sqrt{ \tan^2{ \left( \theta \right) } + 1 } \, d\theta } \\ &= \frac{1}{8} \int{ \tan^2{\left( \theta \right) } \sec^2{ \left( \theta \right) } \sec{\left( \theta \right) } \,d\theta } \\ &= \frac{1}{8} \int{ \left[ \sec^2{\left( \theta \right) } - 1 \right] \sec^3{ \left( \theta \right) } \,d\theta } \\ &= \frac{1}{8} \int{ \sec^5{ \left( \theta \right) } - \sec^3{ \left( \theta \right) } \, d\theta } \\ &= \frac{1}{8}\int{ \frac{1}{\cos^5{ \left( \theta \right) } } - \frac{1}{\cos^3{ \left( \theta \right) } } \, d\theta } \\ &= \frac{1}{8} \int{ \frac{\cos{ \left( \theta \right) } }{\cos^6{ \left( \theta \right) } } - \frac{\cos{ \left( \theta \right) } }{\cos^4{ \left( \theta \right) } } \, d\theta } \\ &= \frac{1}{8} \int{ \cos{ \left( \theta \right) } \left\{ \frac{1}{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^3 } - \frac{ 1}{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^2 } \right\} \, d\theta } \end{align*}[/tex]
Now let [tex]\displaystyle \begin{align*} u = \sin{ \left( \theta \right) } \implies du = \cos{ \left( \theta \right) } \, d\theta \end{align*}[/tex] and the integral becomes
[tex]\displaystyle \begin{align*} \frac{1}{8} \int{ \cos{ \left( \theta \right) } \left\{ \frac{1}{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^3} - \frac{1}{ \left[ 1 - \sin^2{ \left( \theta \right) } \right] ^2 } \right\} \,d\theta } &= \frac{1}{8} \int{ \frac{1}{ \left( 1 - u^2 \right) ^3 } - \frac{1}{ \left( 1 - u^2 \right) ^2 } \, du } \\ &= \frac{1}{8} \int{ \frac{1}{ \left( 1 - u \right) ^3 \left( 1 + u \right) ^3 } - \frac{1}{ \left( 1 - u \right) ^2 \left( 1 + u \right) ^2 } \, du } \end{align*}[/tex]
Which can be solved using partial fractions.
Hmmm, that was a little too messy for my taste. Maybe a different original substitution would have been easier. Try [tex]\displaystyle \begin{align*} x + 1 = \frac{1}{2}\sinh{ (t) } \implies dx = \frac{1}{2}\cosh{(t)}\,dt \end{align*}[/tex] and the integral becomes
[tex]\displaystyle \begin{align*} \int{ \left( x + 1 \right) ^2 \, \sqrt{4 \left( x + 1 \right) ^2 + 1 } \, dx } &= \int{ \left[ \frac{1}{2} \sinh{(t)} \right] ^2 \sqrt{ 4 \left[ \frac{1}{2}\sinh{(t)} \right] ^2 + 1 } \, \frac{1}{2}\cosh{(t)}\,dt } \\ &= \frac{1}{8} \int{ \sinh^2{(t)}\cosh{(t)} \, \sqrt{ \sinh^2{(t)} + 1 } \, dt } \\ &= \frac{1}{8} \int{ \sinh^2{(t)}\cosh^2{(t)} \, dt } \\ &= \frac{1}{8} \int{ \left[ \frac{1}{2} \sinh{(2t)} \right] ^2 \, dt } \\ &= \frac{1}{32} \int{ \sinh^2{(2t)} \, dt } \\ &= \frac{1}{32} \int{ \frac{1}{2} \left[ \cosh{(4t)} - 1 \right] \, dt } \\ &= \frac{1}{64} \int{ \cosh{(4t)} - 1 \, dt } \\ &= \frac{1}{64} \left[ \frac{1}{4}\sinh{(4t)} - t \right] + C \\ &= \frac{1}{256}\sinh{(4t)} - \frac{1}{64}t + C \\ &= \frac{1}{128}\sinh{(2t)}\cosh{(2t)} - \frac{1}{64}t + C \\ &= \frac{1}{64}\sinh{(t)}\cosh{(t)} \left[ 2 \sinh^2{(t)} + 1 \right] - \frac{1}{64}t + C \\ &= \frac{1}{64} \sinh{(t)} \left[ 2 \sinh^2{(t)} + 1 \right] \, \sqrt{ \sinh^2{(t)} + 1 } - \frac{1}{64}t + C \\ &= \frac{1}{32} \left( x + 1 \right) \left\{ 2 \left[ 2 \left( x + 1 \right) \right] ^2 + 1 \right\} \, \sqrt{ \left[ 2 \left( x + 1 \right) \right] ^2 + 1 } - \frac{1}{64}\,\textrm{arsinh}\, { \left[ 2 \left( x + 1 \right) \right] } + C \\ &= \frac{1}{32} \left( x + 1 \right) \left[ 8 \left( x + 1 \right) ^2 + 1 \right] \, \sqrt{ 4 \left( x + 1 \right) ^2 + 1 } - \frac{1}{64} \, \textrm{ arsinh } \, { \left[ 2 \left( x + 1 \right) \right] } + C \end{align*}[/tex]
PHEW!
You have [tex]\displaystyle \begin{align*} \int{ \left( x + 1 \right) ^2 \sqrt{ 4 \left( x + 1 \right) ^2 + 1 } \, dx } \end{align*}[/tex], I would lean towards doing a substitution here...
[tex]\displaystyle \begin{align*} x + 1 = \frac{1}{2}\tan{\left( \theta \right) } \implies dx = \frac{1}{2} \sec^2{ \left( \theta \right) } \, d\theta \end{align*}[/tex] and the integral becomes
[tex]\displaystyle \begin{align*} \int{ \left( x + 1 \right) ^2 \sqrt{4 \left( x + 1 \right) ^2 + 1 } \, dx} &= \int{ \left[ \frac{1}{2}\tan{ \left( \theta \right) } \right] ^2 \sqrt{ 4 \left[ \frac{1}{2} \tan{ \left( \theta \right) } \right] ^2 + 1 } \, \frac{1}{2} \sec^2{ \left( \theta \right) } \, d\theta } \\ &= \frac{1}{2} \int{ \frac{1}{4}\tan^2{ \left( \theta \right) } \sec^2{ \left( \theta \right) } \, \sqrt{ \tan^2{ \left( \theta \right) } + 1 } \, d\theta } \\ &= \frac{1}{8} \int{ \tan^2{\left( \theta \right) } \sec^2{ \left( \theta \right) } \sec{\left( \theta \right) } \,d\theta } \\ &= \frac{1}{8} \int{ \left[ \sec^2{\left( \theta \right) } - 1 \right] \sec^3{ \left( \theta \right) } \,d\theta } \\ &= \frac{1}{8} \int{ \sec^5{ \left( \theta \right) } - \sec^3{ \left( \theta \right) } \, d\theta } \\ &= \frac{1}{8}\int{ \frac{1}{\cos^5{ \left( \theta \right) } } - \frac{1}{\cos^3{ \left( \theta \right) } } \, d\theta } \\ &= \frac{1}{8} \int{ \frac{\cos{ \left( \theta \right) } }{\cos^6{ \left( \theta \right) } } - \frac{\cos{ \left( \theta \right) } }{\cos^4{ \left( \theta \right) } } \, d\theta } \\ &= \frac{1}{8} \int{ \cos{ \left( \theta \right) } \left\{ \frac{1}{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^3 } - \frac{ 1}{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^2 } \right\} \, d\theta } \end{align*}[/tex]
Now let [tex]\displaystyle \begin{align*} u = \sin{ \left( \theta \right) } \implies du = \cos{ \left( \theta \right) } \, d\theta \end{align*}[/tex] and the integral becomes
[tex]\displaystyle \begin{align*} \frac{1}{8} \int{ \cos{ \left( \theta \right) } \left\{ \frac{1}{\left[ 1 - \sin^2{ \left( \theta \right) } \right] ^3} - \frac{1}{ \left[ 1 - \sin^2{ \left( \theta \right) } \right] ^2 } \right\} \,d\theta } &= \frac{1}{8} \int{ \frac{1}{ \left( 1 - u^2 \right) ^3 } - \frac{1}{ \left( 1 - u^2 \right) ^2 } \, du } \\ &= \frac{1}{8} \int{ \frac{1}{ \left( 1 - u \right) ^3 \left( 1 + u \right) ^3 } - \frac{1}{ \left( 1 - u \right) ^2 \left( 1 + u \right) ^2 } \, du } \end{align*}[/tex]
Which can be solved using partial fractions.
Hmmm, that was a little too messy for my taste. Maybe a different original substitution would have been easier. Try [tex]\displaystyle \begin{align*} x + 1 = \frac{1}{2}\sinh{ (t) } \implies dx = \frac{1}{2}\cosh{(t)}\,dt \end{align*}[/tex] and the integral becomes
[tex]\displaystyle \begin{align*} \int{ \left( x + 1 \right) ^2 \, \sqrt{4 \left( x + 1 \right) ^2 + 1 } \, dx } &= \int{ \left[ \frac{1}{2} \sinh{(t)} \right] ^2 \sqrt{ 4 \left[ \frac{1}{2}\sinh{(t)} \right] ^2 + 1 } \, \frac{1}{2}\cosh{(t)}\,dt } \\ &= \frac{1}{8} \int{ \sinh^2{(t)}\cosh{(t)} \, \sqrt{ \sinh^2{(t)} + 1 } \, dt } \\ &= \frac{1}{8} \int{ \sinh^2{(t)}\cosh^2{(t)} \, dt } \\ &= \frac{1}{8} \int{ \left[ \frac{1}{2} \sinh{(2t)} \right] ^2 \, dt } \\ &= \frac{1}{32} \int{ \sinh^2{(2t)} \, dt } \\ &= \frac{1}{32} \int{ \frac{1}{2} \left[ \cosh{(4t)} - 1 \right] \, dt } \\ &= \frac{1}{64} \int{ \cosh{(4t)} - 1 \, dt } \\ &= \frac{1}{64} \left[ \frac{1}{4}\sinh{(4t)} - t \right] + C \\ &= \frac{1}{256}\sinh{(4t)} - \frac{1}{64}t + C \\ &= \frac{1}{128}\sinh{(2t)}\cosh{(2t)} - \frac{1}{64}t + C \\ &= \frac{1}{64}\sinh{(t)}\cosh{(t)} \left[ 2 \sinh^2{(t)} + 1 \right] - \frac{1}{64}t + C \\ &= \frac{1}{64} \sinh{(t)} \left[ 2 \sinh^2{(t)} + 1 \right] \, \sqrt{ \sinh^2{(t)} + 1 } - \frac{1}{64}t + C \\ &= \frac{1}{32} \left( x + 1 \right) \left\{ 2 \left[ 2 \left( x + 1 \right) \right] ^2 + 1 \right\} \, \sqrt{ \left[ 2 \left( x + 1 \right) \right] ^2 + 1 } - \frac{1}{64}\,\textrm{arsinh}\, { \left[ 2 \left( x + 1 \right) \right] } + C \\ &= \frac{1}{32} \left( x + 1 \right) \left[ 8 \left( x + 1 \right) ^2 + 1 \right] \, \sqrt{ 4 \left( x + 1 \right) ^2 + 1 } - \frac{1}{64} \, \textrm{ arsinh } \, { \left[ 2 \left( x + 1 \right) \right] } + C \end{align*}[/tex]
PHEW!
Hello, Yuuki!
. . [tex]4(x^2+2x+1) + 1 \:=\:4(x+1)^2+1[/tex]
The integral becomes: .[tex]\int (x+1)^2\sqrt{4(x+1)^2+1}\,dx[/tex]
Let [tex]u \,=\,x+1\quad\Rightarrow\quad dx \,=\,du[/tex]
Then we have: .[tex]\int u^2\sqrt{4u^2+1}\,du[/tex]
Let [tex]u \,=\,\tfrac{1}{2}\tan\theta \quad \Rightarrow\quad du \,=\,\tfrac{1}{2}\sec^2\!\theta\,d\theta[/tex]
And we have: .[tex]\int\left(\tfrac{1}{2}\tan\theta\right)^2\left( \sec\theta\right)\left( \tfrac{1}{2}\sec^2\!\theta\,d\theta\right)[/tex]
. . [tex]=\;\tfrac{1}{8}\int\sec^3\!\theta\tan^2\!\theta\,d\theta \;=\;\tfrac{1}{8}\int\sec^3\!\theta(\sec^2\!\theta -1)\,d\theta [/tex]
. . [tex]=\;\tfrac{1}{8}\int(\sec^5\!\theta - \sec^3\!\theta)\,d\theta[/tex]
Apply the reduction formula:
. . [tex]\int \sec^n\!x\,dx \;=\;\frac{1}{2}\left[\sec^{n-2}\!x\tan x + \int\sec^{n-2}\!x\,dx\right][/tex]
and remember to back-substitute.
Under the radical, we have:
. . [tex]4(x^2+2x+1) + 1 \:=\:4(x+1)^2+1[/tex]
The integral becomes: .[tex]\int (x+1)^2\sqrt{4(x+1)^2+1}\,dx[/tex]
Let [tex]u \,=\,x+1\quad\Rightarrow\quad dx \,=\,du[/tex]
Then we have: .[tex]\int u^2\sqrt{4u^2+1}\,du[/tex]
Let [tex]u \,=\,\tfrac{1}{2}\tan\theta \quad \Rightarrow\quad du \,=\,\tfrac{1}{2}\sec^2\!\theta\,d\theta[/tex]
And we have: .[tex]\int\left(\tfrac{1}{2}\tan\theta\right)^2\left( \sec\theta\right)\left( \tfrac{1}{2}\sec^2\!\theta\,d\theta\right)[/tex]
. . [tex]=\;\tfrac{1}{8}\int\sec^3\!\theta\tan^2\!\theta\,d\theta \;=\;\tfrac{1}{8}\int\sec^3\!\theta(\sec^2\!\theta -1)\,d\theta [/tex]
. . [tex]=\;\tfrac{1}{8}\int(\sec^5\!\theta - \sec^3\!\theta)\,d\theta[/tex]
Apply the reduction formula:
. . [tex]\int \sec^n\!x\,dx \;=\;\frac{1}{2}\left[\sec^{n-2}\!x\tan x + \int\sec^{n-2}\!x\,dx\right][/tex]
and remember to back-substitute.