# trig substitution

#### softballchick

there is no numerator... now i'm confused. i know i use the trig sub is x=a tan theta

#### 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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#### softballchick

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
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

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

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

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

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

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

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

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

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

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