# trig substitution

• Aug 4th 2010, 09:24 PM
softballchick
trig substitution
Attachment 18445

there is no numerator... now i'm confused. i know i use the trig sub is x=a tan theta
• Aug 4th 2010, 09:29 PM
Also sprach Zarathustra
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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• Aug 4th 2010, 10:13 PM
softballchick
ok so I get

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

t=sec(k)
so dt is the derivative of sec?
• Aug 5th 2010, 12:33 AM
Prove It
Hyperbolic substitution is easier in this case than trigonometric...

$\int{\sqrt{x^2 + 2x}\,dx} = \int{\sqrt{x^2 + 2x + 1^2 - 1^2}\,dx}$

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

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

$\int{\sqrt{\cosh^2{t} - 1}\,\sinh{t}\,dt}$

$= \int{\sqrt{\sinh^2{t}}\,\sinh{t}\,dt}$

$= \int{\sinh{t}\sinh{t}\,dt}$

$= \int{\sinh^2{t}\,dt}$

$= \int{\frac{\cosh{2t}}{2} - \frac{1}{2}\,dt}$

$= \frac{\sinh{2t}}{4} - \frac{t}{2} + C$

$= \frac{2\cosh{t}\sqrt{\cosh^2{t} - 1}}{2} - \frac{t}{2} + C$

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

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