# Trig Substitution

• Oct 13th 2010, 08:52 PM
andrewho93
Trig Substitution
Integrate (√(1+x^2))/(x))

I get to the integral of 1/(sinxcos²x)
What would I do next
• Oct 13th 2010, 09:11 PM
yannyy
use u=cosx and then do another trig substitution if you get stuck.
• Oct 13th 2010, 09:50 PM
TheCoffeeMachine
Quote:

Originally Posted by andrewho93
Integrate (√(1+x^2))/(x))

Without a trigonometric substitution, doing in this way yields a very nice answer:

$\displaystyle u = \sqrt{1+x^2} \Leftrightarrow dx = \frac{\sqrt{1+x^2}}{x}\;{du}$ $\displaystyle \Rightarrow \int\frac{\sqrt{1+x^2}}{x}\;{dx} = \int\left(\frac{\sqrt{1+x^2}}{x}\right)\left(\frac {\sqrt{1+x^2}}{x}\right)\;{du} = \int\frac{u^2}{u^2-1}\;{du}$ $\displaystyle = \int\frac{u^2}{(u+1)(u-1)}\;{du} = \int \left\{\frac{1}{2(u-1)}-\frac{1}{2(u+1)}+1 \right\}\;{du}[/tex] [LaTeX ERROR: Compile failed] " alt=" \displaystyle = \int\frac{u^2}{(u+1)(u-1)}\;{du} = \int \left\{\frac{1}{2(u-1)}-\frac{1}{2(u+1)}+1 \right\}\;{du}[/tex] [LaTeX ERROR: Compile failed] " />[LaTeX ERROR: Compile failed] $[tex]\displaystyle = \boxed{\ln\sqrt{\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+\sqrt{1+x^2}+k}.$

Quote:

I get to the integral of 1/(sinxcos²x)
What would I do next
Hint: [LaTeX ERROR: Compile failed] .
• Oct 13th 2010, 09:55 PM
yannyy
oh that's a very smart hint
• Oct 14th 2010, 07:09 AM
Krizalid
i like when trig. is not everything, here's another way to tackle it:

consider $\displaystyle\int{\frac{dx}{x\sqrt{1+{{x}^{2}}}}},$ substitute $x=\dfrac1t$ and the integral becomes $-\displaystyle\frac{dt}{\sqrt{1+t^2}}=-\ln\big(t+\sqrt{1+t^2}\big)+c_1,$ back substitute and get $\displaystyle\int{\frac{dx}{x\sqrt{1+{{x}^{2}}}}}= \ln \left( \frac{x}{1+\sqrt{1+{{x}^{2}}}} \right)+\sqrt{1+{{x}^{2}}}+k.$