integral with sqrt in denominator

**Boo** · January 27th 2013, 02:33 AM

Hello!
Please, what should I do with the sqrt in the denominator?

$\int \frac{dx}{\sqrt{2+3x-2x^2}}$
Please, help someone!!!!
Many thanks!

**ibdutt** · January 27th 2013, 03:27 AM

Workout the denominator in the form ∫▒〖1/√(a^2-x^2 ) OR 〗 ∫▒〖1/√(a^2+x^2 ) 〗and then use the formula
∫▒〖1/√(a^2-x^2 ) dx = sin^(-1)⁡〖x/a+C OR 〗 ∫▒〖1/√(a^2+x^2 ) 〗 dx= log⁡〖|x+ 〗 √(a^2+x^2 )|+C〗
2 + 3x – 2x2 = 2 – 2[ x2 - 3/2 x] = 2 – 2[ x2 - 3/2 x + 〖(3/4)〗^2- 〖(3/4)〗^2]
= 2 – 2[ 〖( x- 3/4)〗^2- 〖(3/4)〗^2]
= 2[ 1 – 〖( x- 3/4)〗^2+ 9/16]
= 2[25/16 – 〖( x- 3/4)〗^2]
Etc…

**ILikeSerena** · January 27th 2013, 03:38 AM

Originally Posted by Boo

Hello!
Please, what should I do with the sqrt in the denominator?

$\int \frac{dx}{\sqrt{2+3x-2x^2}}$
Please, help someone!!!!
Many thanks!

Well, let's see if we can find a known anti-derivative for an expression with a square root in the denominator.
Hmm, we know that the anti-derivative of $1 \over \sqrt{1 - u^2}$ is $\arcsin u$ .
This looks a bit like it.

Can you perhaps write the argument of the square root in the form $a(1-(bx-c)^2)$ ?

After that you can make the substitution u=bx-c.

**Boo** · January 27th 2013, 04:11 AM

Thank u, both!!!
Well, 2+3x-2x^2=1-(bx-c)^2
1+3x-2x^2=-(bx-c)^2= - (b^2x^2-2bc+c^2)
=-b^2+2bc-c^2
dont see the soution: -c^2should be 1 then...

**Prove It** · January 27th 2013, 04:18 AM

Originally Posted by Boo

Hello!
Please, what should I do with the sqrt in the denominator?

$\int \frac{dx}{\sqrt{2+3x-2x^2}}$
Please, help someone!!!!
Many thanks!

$\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{2 + 3x - 2x^2}}} &= \int{\frac{dx}{\sqrt{-2\left( x^2 - \frac{3}{2}x - 1 \right)}}} \\ &= \int{\frac{dx}{\sqrt{-2\left[ x^2 - \frac{3}{2}x + \left( -\frac{3}{4} \right)^2 - \left( -\frac{3}{4} \right)^2 - 1 \right]}}} \\ &= \int{\frac{dx}{\sqrt{-2 \left[ \left( x - \frac{3}{4} \right)^2 - \frac{25}{16} \right] }}} \\ &= \frac{1}{\sqrt{2}} \int{ \frac{dx}{\sqrt{\frac{25}{16} - \left( x - \frac{3}{4} \right)^2 }} } \end{align*}$

Now make the substitution $\displaystyle \begin{align*} x - \frac{3}{4} = \frac{5}{4}\sin{\theta} \implies dx = \frac{5}{4}\cos{\theta}\,d\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{\sqrt{2}} \int{ \frac{dx}{\sqrt{ \frac{25}{16} - \left( x - \frac{3}{4} \right)^2 }} } &= \frac{1}{\sqrt{2}} \int{ \frac{ \frac{5}{4}\cos{\theta}\,d\theta }{ \sqrt{ \frac{25}{16} - \left( \frac{5}{4}\sin{\theta} \right)^2 } } } \\ &= \frac{1}{\sqrt{2}}\int{ \frac{ \frac{5}{4}\cos{\theta}\,d\theta }{ \sqrt{ \frac{25}{16} - \frac{25}{16}\sin^2{\theta} } } } \\ &= \frac{1}{\sqrt{2}} \int{ \frac{\frac{5}{4}\cos{\theta}\,d\theta}{ \sqrt{ \frac{25}{16} \left( 1 - \sin^2{\theta} \right) } } } \\ &= \frac{1}{\sqrt{2}} \int{ \frac{ \frac{5}{4}\cos{\theta}\,d\theta }{ \sqrt{ \frac{25}{16}\cos^2{\theta} } } } \\ &= \frac{1}{\sqrt{2}} \int{ \frac{ \frac{5}{4}\cos{\theta}\,d\theta }{ \frac{5}{4}\cos{\theta} } } \\ &= \frac{1}{\sqrt{2}} \int{1\,d\theta} \\ &= \frac{1}{\sqrt{2}} \theta + C \\ &= \frac{1}{\sqrt{2}} \arcsin{ \left[ \frac{4}{5} \left( x - \frac{3}{4} \right) \right] } + C \end{align*}$

**Boo** · January 27th 2013, 04:21 AM

I shoudl solve it like this:

2+3x-2x^2= (25/8) (1-(x-3/4)^2/(25/16)

and put in:

t= 4(x-3/4)/5

huh...any help???

**ILikeSerena** · January 27th 2013, 04:32 AM

You have left out the factor a...

But let's see:

$2+3x-2x^2$

$\quad = -2(x^2 -\frac 32 x - 1)$

$\quad = -2((x-\frac 34)^2 - \frac 9{16} - 1)$

$\quad = +2(\frac{25}{16} - (x - \frac 34)^2)$

$\quad = +2 \cdot \frac{25}{16}(1 - \frac{16}{25}(x - \frac 34)^2)$

$\quad = \frac{25}{8}(1 - (\frac 45 x - \frac 35)^2)$

So... substitute:

$u = \frac 45 x - \frac 35$

**Boo** · January 28th 2013, 06:25 AM

Please how did u get:

$x-\frac{3}{4}=\frac{5}{4}sin\alpha$

**ILikeSerena** · January 28th 2013, 10:25 AM

Originally Posted by Boo

Please how did u get:

$x-\frac{3}{4}=\frac{5}{4}sin\alpha$

Can you substitute it?
If you do, I think your question will be answered.
It's a smart choice for the substitution.

**Prove It** · January 28th 2013, 02:37 PM

Originally Posted by Boo

Please how did u get:

$x-\frac{3}{4}=\frac{5}{4}sin\alpha$

Experience. Basically when you see something of the form $\displaystyle \begin{align*} \sqrt{ a^2 - \left( x - b \right)^2 } \end{align*}$ in the denominator and you can't do a u-substitution, then a substitution of $\displaystyle \begin{align*} x - b = a\sin{\theta} \end{align*}$ is appropriate, as it will simplify using the Pythagorean Identity and then cancel with its own derivative of $\displaystyle \begin{align*} a\cos{\theta} \end{align*}$ , making the integration unbelievably easy.

**ILikeSerena** · January 28th 2013, 02:49 PM

Originally Posted by Prove It

Experience. Basically when you see something of the form $\displaystyle \begin{align*} \sqrt{ a^2 - \left( x - b \right)^2 } \end{align*}$ in the denominator and you can't do a u-substitution, then a substitution of $\displaystyle \begin{align*} x - b = a\sin{\theta} \end{align*}$ is appropriate, as it will simplify using the Pythagorean Identity and then cancel with its own derivative of $\displaystyle \begin{align*} a\cos{\theta} \end{align*}$ , making the integration unbelievably easy.

As an alternative I propose:

$\int {dx \over \sqrt{2+3x-2x^2}} = ...$

$= \int {dx \over \sqrt{\frac{25}{8}(1-(\frac 45 x - \frac 35)^2)}}$

$= \sqrt{\frac{8}{25}} \int {dx \over \sqrt{1-(\frac 45 x - \frac 35)^2}}$

$= \sqrt{\frac{8}{25}} \cdot \frac 54 \arcsin(\frac 45 x - \frac 35) + C$

$= {1 \over \sqrt{2}} \arcsin(\frac 45 x - \frac 35) + C$

Why introduce a sine, or a substitution for that matter, when you can get straight to the answer?

**Prove It** · January 28th 2013, 02:52 PM

Originally Posted by ILikeSerena

As an alternative I propose:

$\int {dx \over \sqrt{2+3x-2x^2}} = ...$

$= \int {dx \over \sqrt{\frac{25}{8}(1-(\frac 45 x - \frac 35)^2)}}$

$= \sqrt{\frac{8}{25}} \int {dx \over \sqrt{1-(\frac 45 x - \frac 35)^2}}$

$= \sqrt{\frac{8}{25}} \cdot \frac 54 \arcsin(\frac 45 x - \frac 35) + C$

$= {1 \over \sqrt{2}} \arcsin(\frac 45 x - \frac 35) + C$

Why introduce a sine, or a substitution for that matter, when you can get straight to the answer?

It depends whether or not you prefer reading a result from tables of integrals, or if you prefer being able to get TO those results from the tables. I think it shows a much deeper understanding of how integration works to be able to do the substitution.

**Boo** · January 29th 2013, 03:13 AM

PLEASE; HOW DID U GET .

$x-b=asin\alpha$ ?
Many Thanks-stuck at basic things!!!

**Prove It** · January 29th 2013, 05:11 AM

Originally Posted by Boo

PLEASE; HOW DID U GET .

$x-b=asin\alpha$ ?
Many Thanks-stuck at basic things!!!

Like I told you, it comes with experience. Let's just take a look at the arbitrary integral $\displaystyle \begin{align*} \int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{a^2 - \left[ a\sin{(\theta)} \right]^2}}} \end{align*}$ .

Notice that it looks extremely complicated. One might think that since there is a sine function in the integrand and the cosine function (sine's derivative) is a multiple, it might be appropriate to do a substitution. So if we let $\displaystyle \begin{align*} x = a\sin{(\theta)} \implies dx = a\cos{(\theta)}\,d\theta \end{align*}$ , then we would have

$\displaystyle \begin{align*} \int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{a^2 - \left[ a\sin{(\theta)} \right]^2 }}} &= \frac{1}{a}\int{\frac{ a\cos{ (\theta) }\,d\theta}{ \sqrt{a^2 - \left[ a\sin{ ( \theta ) } \right]^2 }}} \\ &= \frac{1}{a}\int{\frac{dx}{\sqrt{a^2 - x^2}}} \end{align*}$

But this does NOT help us (in fact, it gets us to almost the exact same integral that we were trying to find in your original post!) so a different tactic is required.

You should be well aware of the Pythagorean Identity in trigonometry: $\displaystyle \begin{align*} \sin^2{(\theta)} + \cos^2{(\theta)} \equiv 1 \end{align*}$ . We can use this fact to help us simplify the integrand. Notice that this means $\displaystyle \begin{align*} \cos^2{(\theta)} \equiv 1 - \sin^2{(\theta)} \end{align*}$ . Can you see that we have almost this exact quantity in the denominator of the starting integral in this post?

$\displaystyle \begin{align*} \int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{ a^2 - \left[ a\sin{(\theta)} \right]^2 }}} &= \int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{ a^2 - a^2\sin^2{(\theta)} }}} \\ &= \int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{a^2 \left[ 1 - \sin^2{(\theta)} \right] }}} \\ &= \int{\frac{\cos{(\theta)}\,d\theta}{\sqrt{a^2 \cos^2{(\theta)}}}} \\ &= \int{ \frac{ \cos{ ( \theta ) } \, d\theta }{ |a| \cos{ ( \theta ) } } } \\ &= \frac{1}{|a|}\int{d\theta} \end{align*}$

which is an EXTREMELY easy quantity to integrate.

So that means that when you have something of the form $\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{a^2 - x^2}}} \end{align*}$ or similar, it is appropriate to use the substitution $\displaystyle \begin{align*} x = a\sin{(\theta)} \end{align*}$ due to the fact that the denominator simplifies to a cosine function by Pythagoras, and you also end up with a cosine function in the derivative which it cancels with, giving you an extremely easy quantity (such as a constant) to integrate.

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