How do you prove the integral formula #intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C# ?
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To prove the integral formula
[ \int \frac{dx}{\sqrt{x^2 + a^2}} = \ln(x + \sqrt{x^2 + a^2}) + C ]
where ( C ) is the constant of integration, you can use a trigonometric substitution. Let ( x = a \sinh(t) ). Then, ( dx = a \cosh(t) , dt ).
Substituting these into the integral:
[ \int \frac{a \cosh(t) , dt}{\sqrt{(a\sinh(t))^2 + a^2}} ]
Simplify:
[ \int \frac{a \cosh(t) , dt}{\sqrt{a^2(\sinh(t))^2 + a^2}} ]
[ = \int \frac{a \cosh(t) , dt}{\sqrt{a^2(\sinh(t))^2 + a^2}} ]
[ = \int \frac{a \cosh(t) , dt}{\sqrt{a^2(\sinh(t))^2 + a^2}} ]
[ = \int \frac{a \cosh(t) , dt}{a\sqrt{(\sinh(t))^2 + 1}} ]
[ = \int \frac{\cosh(t) , dt}{\sqrt{(\sinh(t))^2 + 1}} ]
Now, note that ( \cosh^2(t) - \sinh^2(t) = 1 ), so ( \sinh^2(t) = \cosh^2(t) - 1 ).
[ = \int \frac{\cosh(t) , dt}{\sqrt{\cosh^2(t) - 1 + 1}} ]
[ = \int \frac{\cosh(t) , dt}{\sqrt{\cosh^2(t)}} ]
[ = \int dt ]
[ = t + C ]
Finally, substitute back ( t ) in terms of ( x ):
[ t = \text{arsinh}\left(\frac{x}{a}\right) ]
[ = \text{arsinh}\left(\frac{x}{a}\right) + C ]
[ = \ln\left(x + \sqrt{x^2 + a^2}\right) + C ]
Thus, the formula is proven.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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