# Thread: Integrals with square roots

1. ## Integrals with square roots

Hi,

I'm studying for a test and I'm doing some integral exercises. I manage to solve a lot of them, but I'm always having difficulty with integrals that have square roots.

I have 4 equations, but I don't want the actual answer to any of them. I would just like a push in the right direction for solving them.

$\displaystyle \int{\sqrt{x}}\arcsin{\sqrt{x}}$
$\displaystyle \int{x\sqrt{L^2-x^2}}$
$\displaystyle \int{\frac{1}{\sqrt{1+e^x}}}$
$\displaystyle \int{\frac{x^2}{\sqrt{4-3x}}}$

Thanks for any help!

2. For the second...

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

Now let $\displaystyle u = L^2 - x^2$ so that $\displaystyle \frac{du}{dx} = -2x$ and the integral becomes

$\displaystyle -\frac{1}{2}\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx }$

$\displaystyle = -\frac{1}{2}\int{u^{\frac{1}{2}}\,du}$

$\displaystyle = -\frac{1}{2}\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2 }}\right) + C$

$\displaystyle = -\frac{1}{2}\left(\frac{2}{3}u^{\frac{3}{2}}\right) +C$

$\displaystyle = -\frac{1}{3}\sqrt{(L^2 - x^2)^3} + C$.

3. Thanks, that was easier than I thought (But you didn't have to give the entire answer )

4. For 1, use parts, integrating root x and differentiating the other, then make a u-sub of the root which is in the denominator of the subsequent integrand.

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

And...

... is the product rule.

The general drift for integration by parts is...

Spoiler:

Margin work to be able to integrate the right function with respect to the root of the difference 1 minus x... i.e. with respect to the dashed balloon function which we normally label u...

$\displaystyle \frac{2}{3}\ x^{\frac{3}{2}}\ \frac{1}{\sqrt{1 - x}}\ \frac{1}{2}\ \frac{1}{\sqrt{x}}$

$\displaystyle \ =\ \frac{1}{3}\ \frac{x}{\sqrt{1 - x}}$

$\displaystyle \ =\ \frac{1}{3}\ \frac{1 - u^2}{\sqrt{1 - x}}$

$\displaystyle \ =\ \frac{2}{3}\ (u^2 - 1)\ (-\frac{1}{2})\ \frac{1}{\sqrt{1 - x}}$

Third one, you can substitute $\displaystyle \tan^2 \theta$ for $\displaystyle e^x$, though I'm not certain you need to.

Number four is very similar to just the latter part of number one.

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

5. For the first, I now have:
$\displaystyle 2x^{3/2}*arcsin(\sqrt{x}) - \frac{1}{3}\int{\frac{v}{\sqrt{1-v}}}$
Where v = uČ and u = sqrt(x)

Which substitution do I need to solve the last integral?

6. You're right except for the first coefficient (see my first pic), and if I'm not wrong then the last integral to do is...

$\displaystyle \frac{2}{3}\ \int (s^2 - 1)\ ds$

Just in case a picture helps with number 3...

... where (key in spoiler) ...

Spoiler:

... the general drift is...

And the rest...

Spoiler:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

7. Oh, I forgot the /3 at the start.

And I can't really deduce from your image what substitution I need to use on $\displaystyle \frac{1}{3}\int{\frac{v}{\sqrt{1-v}}}$

8. The substitution to use here is u (or s, or my dashed balloon) equals... $\displaystyle \sqrt{1 - v}$ because that will do as the inner function of a chain rule differentiation that we need to work backwards through.

I put this...

$\displaystyle \frac{2}{3}\ x^{\frac{3}{2}}\ \frac{1}{\sqrt{1 - x}}\ \frac{1}{2}\ \frac{1}{\sqrt{x}}$

$\displaystyle \ =\ \frac{1}{3}\ \frac{x}{\sqrt{1 - x}}$

$\displaystyle \ =\ \frac{1}{3}\ \frac{1 - u^2}{\sqrt{1 - x}}$

$\displaystyle \ =\ \frac{2}{3}\ (u^2 - 1)\ (-\frac{1}{2})\ \frac{1}{\sqrt{1 - x}}$

in the spoiler - not sure if you saw that (I did edit it in a bit later). It doesn't contain any differentials, but hopefully you can see that we've expressed x (your v) in terms of u (my dashed balloon). That's how we know that the function to integrate with respect to whatever we're calling the inner function (u or the dashed balloon or s) is

$\displaystyle \ \frac{2}{3}\ (u^2 - 1)$

9. Hello, superdl!

Here's the 4th one . . .

$\displaystyle \displaystyle \int \frac{x^2}{\sqrt{4-3x}}\,dx$

If there is a linear function under the radical,
. . let $\displaystyle u$ equal the entire radical.

Let $\displaystyle u \:=\:\sqrt{4-3x} \quad\Rightarrow\quad x \:=\:\frac{4-u^2}{3} \quad\Rightarrow\quad dx \:=\:-\tfrac{2}{3}u\,du$

$\displaystyle \displaystyle \text{Substitute: }\;\int\frac{\left(\frac{4-u^2}{3}\right)^2}{u}\left(-\tfrac{2}{3}u\,du\right) \;=\;-\tfrac{2}{27}\int(16-8u^2+u^4)\,du$

. . . . . . . $\displaystyle =\;-\tfrac{2}{27}\left(16u - \tfrac {8}{3}u^3 + \tfrac{1}{5}u^5\right) + C$

At this point, you know how to back-substitute, right?
But most textbooks have the answer simplified beyond all recognition.
I'd like to demonstrate how that is done.

We have: .$\displaystyle -\frac{2}{27}\left(16u - \tfrac{8}{3}u^3 + \tfrac{1}{5}u^5\right) + C$

Factor: . $\displaystyle -\dfrac{2}{27}\cdot\dfrac{u}{15}\left(240 - 40u^2 + 3u^4\right) + C$

Back-substitute: .$\displaystyle -\dfrac{2}{405}\sqrt{4-3x}\,\bigg[240 - 40(4-3x) + 3(4-3x)^2\bigg] + C$

. . which simplifies to: .$\displaystyle -\dfrac{2}{405}\sqrt{4-3x}\,\left(27x^2 + 48x + 128\right)+C$

10. Thanks! I get it

So 1, 2 and 4 are solved.

Any suggestions on 3 ?

11. Originally Posted by superdl
[snip]
$\displaystyle \int{\frac{1}{\sqrt{1+e^x}}}$
[snip]
The obvious substitution is u = e^x. Now see here: integrate 1&#47;&#40;x Sqrt&#91;x &#43; 1&#93;&#41; - Wolfram|Alpha(Click on Show steps).

12. One way of doing the first one would be to let $\displaystyle t = \arcsin{\sqrt{x}}$ to get $\displaystyle 2\int{t\sin^2{t}\cos^2{t}\;{dt}$ and then by parts:

$\displaystyle \displaystyle \frac{1}{8}\int t-t\cos{4t}\;{dt} = \frac{1}{16}t^2-\frac{1}{8}\int{t\cos{4t}}\;{dt} = \frac{1}{16}t^2-\frac{1}{32}t\sin{4t}+\frac{1}{32}\int{\sin{4t}}\: {dt} = \frac{1}{16}t^2-\frac{1}{32}t\sin{4t}-\frac{1}{128}\cos{4t}+k$ $\displaystyle = \frac{1}{16}\left(\arcsin^2{\sqrt{x}}\right)-\frac{1}{32}\left(\arcsin{\sqrt{x}}\right)\sin\lef t(4\arcsin{\sqrt{x}}\right)-\frac{1}{128}\cos\left(4\arcsin{\sqrt{x}}\right)+k$

13. Originally Posted by mr fantastic
The obvious substitution is u = e^x. Now see here: integrate 1&#47;&#40;x Sqrt&#91;x &#43; 1&#93;&#41; - Wolfram|Alpha(Click on Show steps).
Or at least it becomes obvious when you rewrite the integral as

$\displaystyle \int{\frac{1}{\sqrt{1 + e^x}}\,dx} = \int{\frac{e^x}{e^x\sqrt{1 + e^x}}\,dx}$.