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Trig Substitution
- Thread starter Yuuki
- Start date
- Jan 31, 2012
- 253
Re: trig substitution
You can do it without trig substitution.
$$ \int \frac{\sqrt{1+x^{2}}}{x} \ dx $$
Let $u^{2} = 1+x^{2}$.
Then
$$\int \frac{\sqrt{1+x^{2}}}{x} \ dx = \int \frac{u^{2}}{x^{2}} \ du = \int \frac{u^{2}}{u^{2}-1} \ du$$
$$ = \int \Big( 1 + \frac{1}{u^{2}-1} \Big) \ du = \int \Big( 1 + \frac{1}{2(u-1)} - \frac{1}{2(u+1)} \Big) \ du$$
$$ = u + \frac{\ln(u-1)}{2} - \frac{\ln(u+1)}{2} + C$$
$$ = \sqrt{1+x^{2}} + \frac{1}{2} \ln (\sqrt{1+x^{2}}-1) - \frac{1}{2} \ln(\sqrt{1+x^{2}}+1) + C $$
You can do it without trig substitution.
$$ \int \frac{\sqrt{1+x^{2}}}{x} \ dx $$
Let $u^{2} = 1+x^{2}$.
Then
$$\int \frac{\sqrt{1+x^{2}}}{x} \ dx = \int \frac{u^{2}}{x^{2}} \ du = \int \frac{u^{2}}{u^{2}-1} \ du$$
$$ = \int \Big( 1 + \frac{1}{u^{2}-1} \Big) \ du = \int \Big( 1 + \frac{1}{2(u-1)} - \frac{1}{2(u+1)} \Big) \ du$$
$$ = u + \frac{\ln(u-1)}{2} - \frac{\ln(u+1)}{2} + C$$
$$ = \sqrt{1+x^{2}} + \frac{1}{2} \ln (\sqrt{1+x^{2}}-1) - \frac{1}{2} \ln(\sqrt{1+x^{2}}+1) + C $$
- Jan 29, 2012
- 1,151
I would be inclined to do this like Random Variable but since you got this far, sec(u)= 1/cos(u) and tan(u)= sin(u)/cos(u) so that [tex]\frac{sec^3(u)}{tan(u)}= \frac{1}{cos^3(u)}\frac{cos(u)}{sin(u)}= \frac{1}{sin(u)cos^2(u)}[/tex]which has sine to an odd power. We can multiply both numerator and denominator by sin(x) to get [tex]\frac{sin(x)}{sin^2(x)cos^2(x)}= \frac{sin(x)}{(1- cos^2(x))cos^2(x)}[/tex]
Let v= cos(x) so that dv= -sin(x)dx and the integrand becomes [tex]\frac{-dv}{(1- v^2)v^2}[/tex] which can be integrated using "partial fractions".
Last edited:
Hello, Yuuki
I found an approach.
I'm sure someone will have a better way.
[tex]\text{Multiply by }\frac{\tan\theta}{\tan\theta}:\;\int\frac{\sec^3 \!\theta\tan\theta}{\tan^2\!\theta}\,d\theta[/tex]
. . [tex]=\;\int\frac{\sec^3\!\theta\tan \theta\,d\theta}{\sec^2\!\theta-1} \;=\;\int\frac{\sec^2\!\theta}{\sec^2\!\theta-1}( \sec\theta\tan\theta\,d\theta) [/tex]
[tex]\text{Let }u \,=\,\sec\theta \quad\Rightarrow\quad du \,=\,\sec\theta \tan\theta\,d\theta[/tex]
[tex]\text{We have: }\:\int \frac{u^2}{u^2-1}\,du \;=\;\int\left(1 + \frac{1}{u^2-1}\right)du[/tex]
. . . . . . [tex]=\;u + \frac{1}{2}\ln\left|\frac{u-1}{u+1}\right|+C[/tex]
Now back-substitute . . .
I found an approach.
I'm sure someone will have a better way.
[tex]\text{Multiply by }\frac{\tan\theta}{\tan\theta}:\;\int\frac{\sec^3 \!\theta\tan\theta}{\tan^2\!\theta}\,d\theta[/tex]
. . [tex]=\;\int\frac{\sec^3\!\theta\tan \theta\,d\theta}{\sec^2\!\theta-1} \;=\;\int\frac{\sec^2\!\theta}{\sec^2\!\theta-1}( \sec\theta\tan\theta\,d\theta) [/tex]
[tex]\text{Let }u \,=\,\sec\theta \quad\Rightarrow\quad du \,=\,\sec\theta \tan\theta\,d\theta[/tex]
[tex]\text{We have: }\:\int \frac{u^2}{u^2-1}\,du \;=\;\int\left(1 + \frac{1}{u^2-1}\right)du[/tex]
. . . . . . [tex]=\;u + \frac{1}{2}\ln\left|\frac{u-1}{u+1}\right|+C[/tex]
Now back-substitute . . .
Krizalid
Active member
- Feb 9, 2012
- 118
Another solution:
[tex]\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = \int {\frac{{1 + {x^2}}}{{x\sqrt {1 + {x^2}} }}\,dx} = \int {\frac{{dx}}{{x\sqrt {1 + {x^2}} }}} + \int {\frac{x}{{\sqrt {1 + {x^2}} }}\,dx} ,[/tex]
second integral is easy, but if we want to avoid partial fractions for the first one, put $x=\dfrac1t$ and the integral becomes
[tex] - \int {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} = - \ln \left| {t + \sqrt {1 + {t^2}} } \right| + {k_1}.[/tex]
Finally,
[tex]\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = - \ln \left| {\frac{1}{x} + \sqrt {1 + \frac{1}{{{x^2}}}} } \right| + \sqrt {1 + {x^2}} + k.[/tex]
[tex]\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = \int {\frac{{1 + {x^2}}}{{x\sqrt {1 + {x^2}} }}\,dx} = \int {\frac{{dx}}{{x\sqrt {1 + {x^2}} }}} + \int {\frac{x}{{\sqrt {1 + {x^2}} }}\,dx} ,[/tex]
second integral is easy, but if we want to avoid partial fractions for the first one, put $x=\dfrac1t$ and the integral becomes
[tex] - \int {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} = - \ln \left| {t + \sqrt {1 + {t^2}} } \right| + {k_1}.[/tex]
Finally,
[tex]\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = - \ln \left| {\frac{1}{x} + \sqrt {1 + \frac{1}{{{x^2}}}} } \right| + \sqrt {1 + {x^2}} + k.[/tex]