trig substitution

Jul 2010
59
0
trig sub.JPG

there is no numerator... now i'm confused. i know i use the trig sub is x=a tan theta
 
Dec 2009
1,506
434
Russia
sqrt(x^2+2x)=sqrt(x^2+2x+1-1)=sqrt((x+1)^2-1)

sub: t=x+1

sqrt(t^2-1)

sub: t=sec(k)
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Jul 2010
59
0
ok so I get


sqrt(sec^2(k)-1)d theta

t=sec(k)
so dt is the derivative of sec?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Hyperbolic substitution is easier in this case than trigonometric...

\(\displaystyle \int{\sqrt{x^2 + 2x}\,dx} = \int{\sqrt{x^2 + 2x + 1^2 - 1^2}\,dx}\)

\(\displaystyle = \int{\sqrt{(x + 1)^2 - 1}\,dx}\).

Now let \(\displaystyle x + 1 = \cosh{t}\) so that \(\displaystyle dx = \sinh{t}\,dt\) and the integral becomes

\(\displaystyle \int{\sqrt{\cosh^2{t} - 1}\,\sinh{t}\,dt}\)

\(\displaystyle = \int{\sqrt{\sinh^2{t}}\,\sinh{t}\,dt}\)

\(\displaystyle = \int{\sinh{t}\sinh{t}\,dt}\)

\(\displaystyle = \int{\sinh^2{t}\,dt}\)

\(\displaystyle = \int{\frac{\cosh{2t}}{2} - \frac{1}{2}\,dt}\)

\(\displaystyle = \frac{\sinh{2t}}{4} - \frac{t}{2} + C\)

\(\displaystyle = \frac{2\cosh{t}\sqrt{\cosh^2{t} - 1}}{2} - \frac{t}{2} + C\)

\(\displaystyle = \frac{2(x + 1)\sqrt{(x + 1)^2 - 1}}{2} - \frac{\cosh^{-1}(x + 1)}{2} + C\)

\(\displaystyle = \frac{(2x + 2)\sqrt{x^2 + 2x} - \cosh^{-1}(x + 1)}{2} + C\).