# Thread: Two easy integrals

1. ## Two easy integrals

Integrate

$f(x)=\sqrt{1-x^2}$ & $h(x)=\sqrt{x^2-1}$

2. Originally Posted by liyi
Integrate

$f(x)=\sqrt{1-x^2}$ & $h(x)=\sqrt{x^2-1}$
Do the substitutions: x=sint for the first one and x=sect for the second.

3. I'm not sure what the user above means but I'd;

Rearrange
$f'(x)=\sqrt{1-x^2}$
$f'(x)=\sqrt{1} - \sqrt{x^2}$
$f'(x)= 1^{1/2} - x$
Intergrate; add one to power, divide by new power and add c:
$f(x)= 2/3^{3/2} - 1 + c$
I don't think that can be simplified

4. Originally Posted by markcantdomaths
I'm not sure what the user above means but I'd;

Rearrange
$f'(x)=\sqrt{1-x^2}$
$f'(x)=\sqrt{1} - \sqrt{x^2}$
In general $\sqrt{a^2 + b^2} \neq a + b$

curvature's substitutions should work nicely.

-Dan

5. Originally Posted by markcantdomaths
Rearrange
$f'(x)=\sqrt{1-x^2}$
$f'(x)=\sqrt{1} - \sqrt{x^2}$
No the second equality is wrong.

6. Originally Posted by liyi
Integrate

$f(x)=\sqrt{1-x^2}$
Define the function $u(x) = x\sqrt {1 - x^2 } \implies u'(x) = \sqrt {1 - x^2 } - \frac{{x^2 }}
{{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
{{\sqrt {1 - x^2 } }}.$

Make-up: $\sqrt {1 - x^2 } = \frac{1}
{2}\left[ {u'(x) + \frac{1}
{{\sqrt {1 - x^2 } }}} \right].$

Integrate $\int {\sqrt {1 - x^2 } \,dx} = \frac{1}
{2}\left( {x\sqrt {1 - x^2 } + \arcsin x} \right) + k.$

Originally Posted by liyi
Integrate

$h(x)=\sqrt{x^2-1}$
Consider the function $\varphi (x) = x\sqrt {x^2 - 1} \implies \varphi '(x) = \sqrt {x^2 - 1} + \frac{{x^2 }}
{{\sqrt {x^2 - 1} }} = 2\sqrt {x^2 - 1} + \frac{1}
{{\sqrt {x^2 - 1} }}.$

Make-up & integrate

$\int {\sqrt {x^2 - 1} \,dx} = \frac{1}
{2}\left( {x\sqrt {x^2 - 1} - \int {\frac{1}
{{\sqrt {x^2 - 1} }}\,dx} } \right)+k_1.$

For the remaining integral, substitute $u=x+\sqrt{x^2-1},$

$\int {\frac{1}
{{\sqrt {x^2 - 1} }}\,dx} = \int {\frac{1}
{u}\,du} = \ln \left| {x + \sqrt {x^2 - 1} } \right| + k_2.$

And we happily get $\int {\sqrt {x^2 - 1} \,dx} = \frac{1}
{2}\left( {x\sqrt {x^2 - 1} - \ln \left| {x + \sqrt {x^2 - 1} } \right|} \right) + k.$

7. Originally Posted by Krizalid
Define the function $u(x) = x\sqrt {1 - x^2 } \implies u'(x) = \sqrt {1 - x^2 } - \frac{{x^2 }}
{{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
{{\sqrt {1 - x^2 } }}.$

Make-up: $\sqrt {1 - x^2 } = \frac{1}
{2}\left[ {u'(x) + \frac{1}
{{\sqrt {1 - x^2 } }}} \right].$

Integrate $\int {\sqrt {1 - x^2 } \,dx} = \frac{1}
{2}\left( {x\sqrt {1 - x^2 } + \arcsin x} \right) + k.$

Consider the function $\varphi (x) = x\sqrt {x^2 - 1} \implies \varphi '(x) = \sqrt {x^2 - 1} + \frac{{x^2 }}
{{\sqrt {x^2 - 1} }} = 2\sqrt {x^2 - 1} + \frac{1}
{{\sqrt {x^2 - 1} }}.$

Make-up & integrate

$\int {\sqrt {x^2 - 1} \,dx} = \frac{1}
{2}\left( {x\sqrt {x^2 - 1} - \int {\frac{1}
{{\sqrt {x^2 - 1} }}\,dx} } \right)+k_1.$

For the remaining integral, substitute $u=x+\sqrt{x^2-1},$

$\int {\frac{1}
{{\sqrt {x^2 - 1} }}\,dx} = \int {\frac{1}
{u}\,du} = \ln \left| {x + \sqrt {x^2 - 1} } \right| + k_2.$

And we happily get $\int {\sqrt {x^2 - 1} \,dx} = \frac{1}
{2}\left( {x\sqrt {x^2 - 1} - \ln \left| {x + \sqrt {x^2 - 1} } \right|} \right) + k.$
I understood everything except this step here:
$\sqrt {1 - x^2 } - \frac{{x^2 }}
{{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
{{\sqrt {1 - x^2 } }}.$

Can you please explain it.

8. Originally Posted by angel.white
I understood everything except this step here:
$\sqrt {1 - x^2 } - \frac{{x^2 }}
{{\sqrt {1 - x^2 } }} = 2\sqrt {1 - x^2 } - \frac{1}
{{\sqrt {1 - x^2 } }}.$

Can you please explain it.
as in explain why it is true?

$\sqrt{1 - x^2} - \frac {x^2}{\sqrt {1 - x^2}} = \sqrt{1 - x^2} - \frac {x^2 + 1 - 1}{\sqrt {1 - x^2}}$

$= \sqrt{1 - x^2} + \frac {1 - x^2}{\sqrt {1 - x^2}} - \frac 1{\sqrt {1 - x^2}}$

$= \sqrt{1 - x^2} + \sqrt{1 - x^2} - \frac 1{\sqrt {1 - x^2}}$

$= 2 \sqrt{1 - x^2} - \frac 1{\sqrt {1 - x^2}}$

9. I see now, that's a smart way of doing business.