1. ## Trig Substitution

Integrate (√(1+x^2))/(x))

I get to the integral of 1/(sinxcos²x)
What would I do next

2. use u=cosx and then do another trig substitution if you get stuck.

3. Originally Posted by andrewho93
Integrate (√(1+x^2))/(x))
Without a trigonometric substitution, doing in this way yields a very nice answer:

$\displaystyle \displaystyle u = \sqrt{1+x^2} \Leftrightarrow dx = \frac{\sqrt{1+x^2}}{x}\;{du}$ $\displaystyle \displaystyle \Rightarrow \int\frac{\sqrt{1+x^2}}{x}\;{dx} = \int\left(\frac{\sqrt{1+x^2}}{x}\right)\left(\frac {\sqrt{1+x^2}}{x}\right)\;{du} = \int\frac{u^2}{u^2-1}\;{du}$ $\displaystyle \displaystyle = \int\frac{u^2}{(u+1)(u-1)}\;{du} = \int \left\{\frac{1}{2(u-1)}-\frac{1}{2(u+1)}+1 \right\}\;{du}[/Math]$\displaystyle \displaystyle = \frac{1}{2}\int\frac{1}{u-1}\;{du}-\frac{1}{2}\int\frac{1}{u+1}}\;{du}+\int\;{du}$$\displaystyle \displaystyle \displaystyle = \frac{1}{2}\ln(u-1)-\frac{1}{2}\ln(u+1)+u+k = \ln\sqrt{u-1}-\ln\sqrt{u+1}+u+k = \ln\sqrt\frac{u-1}{u+1}+u+k$$$\displaystyle [Math]\displaystyle = \boxed{\ln\sqrt{\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+\sqrt{1+x^2}+k}.$

I get to the integral of 1/(sinxcos²x)
What would I do next
Hint: $\displaystyle 1 = \cos^2{x}+\sin^2{x}$.

4. oh that's a very smart hint

5. i like when trig. is not everything, here's another way to tackle it:

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