The Notes: `int sqrt(1 + x^2) / x dx` Evaluate the integral

Sunday, August 22, 2010

$\displaystyle\int\frac{\sqrt{{{1}+{{x}}^{{2}}}}}{{x}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int\frac{\sqrt{{{1}+{{x}}^{{2}}}}}{{x}}{\left.{d}{x}\right.}$

Let's evaluate using the trigonometric substitution,

Let $\displaystyle{x}={\tan{{\left(\theta\right)}}}$

$\displaystyle{\left.{d}{x}\right.}={{\sec}}^{{2}}{\left(\theta\right)}{d}\theta$

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

Now use the identity: $\displaystyle{1}+{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int\frac{\sqrt{{{{\sec}}^{{2}}{\left(\theta\right)}}}}{{\tan{{\left(\theta\right)}}}}\cdot{{\sec}}^{{2}}{\left(\theta\right)}{d}\theta$

$\displaystyle=\int\frac{{\sec{{\left(\theta\right)}}}}{{\tan{{\left(\theta\right)}}}}\cdot{{\sec}}^{{2}}{\left(\theta\right)}{d}\theta$

$\displaystyle=\int\frac{{{\sec{{\left(\theta\right)}}}{\left({1}+{{\tan}}^{{2}}{\left(\theta\right)}\right)}}}{{\tan{{\left(\theta\right)}}}}{d}\theta$

$\displaystyle=\int{\left(\frac{{{\sec{{\left(\theta\right)}}}}}{{\tan{{\left(\theta\right)}}}}+\frac{{{\sec{{\left(\theta\right)}}}{{\tan}}^{{2}}{\left(\theta\right)}}}{{\tan{{\left(\theta\right)}}}}\right)}{d}\theta$

$\displaystyle=\int\frac{{\sec{{\left(\theta\right)}}}}{{\tan{{\left(\theta\right)}}}}{d}\theta+\int{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\int{\left(\frac{{1}}{{\cos{{\left(\theta\right)}}}}\right)}\cdot{\left(\frac{{1}}{{\frac{{\sin{{\left(\theta\right)}}}}{{\cos{{\left(\theta\right)}}}}}}\right)}{d}\theta+\int{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\int\frac{{1}}{{\sin{{\left(\theta\right)}}}}{d}\theta+\int{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\int{\csc{{\left(\theta\right)}}}{d}\theta+\int{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}{d}\theta$

Now using the standard integrals,

$\displaystyle\int{\csc{{\left({x}\right)}}}{\left.{d}{x}\right.}={\ln}{\mid}{\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}$

$\displaystyle\int{\sec{{\left({x}\right)}}}{\tan{{\left({x}\right)}}}{\left.{d}{x}\right.}={\sec{{\left({x}\right)}}}$

$\displaystyle={\ln}{\left|{\csc{{\left(\theta\right)}}}-{\cot{{\left(\theta\right)}}}\right|}+{\sec{{\left(\theta\right)}}}$

Now substitute back $\displaystyle{x}={\tan{{\left(\theta\right)}}}$

$\displaystyle\Rightarrow{\cot{{\left(\theta\right)}}}=\frac{{1}}{{\tan{{\left(\theta\right)}}}}=\frac{{1}}{{x}}$

$\displaystyle{1}+{{\tan}}^{{2}}{\left(\theta\right)}={{\sec}}^{{2}}{\left(\theta\right)}$

$\displaystyle\Rightarrow{1}+{{x}}^{{2}}={{\sec}}^{{2}}{\left(\theta\right)}$

$\displaystyle\Rightarrow{\sec{{\left(\theta\right)}}}=\sqrt{{{1}+{{x}}^{{2}}}}$

$\displaystyle{1}+{{\cot}}^{{2}}{\left(\theta\right)}={{\csc}}^{{2}}{\left(\theta\right)}$

$\displaystyle\Rightarrow{1}+{{\left(\frac{{1}}{{x}}\right)}}^{{2}}={{\csc}}^{{2}}{\left(\theta\right)}$

$\displaystyle{\csc{{\left(\theta\right)}}}=\frac{\sqrt{{{1}+{{x}}^{{2}}}}}{{x}}$

$\displaystyle\therefore\int\frac{\sqrt{{{1}+{{x}}^{{2}}}}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|\frac{\sqrt{{{1}+{{x}}^{{2}}}}}{{x}}-\frac{{1}}{{x}}\right|}+\sqrt{{{1}+{{x}}^{{2}}}}+{C}$ ,C is a constant

The Notes

Sunday, August 22, 2010

$\displaystyle\int\frac{\sqrt{{{1}+{{x}}^{{2}}}}}{{x}}{\left.{d}{x}\right.}$ Evaluate the integral

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