Approriate u substituion help

[nickman]

The question is on integration, a substitution needs to be made before I can use the back of my book to solve it but I'm confused on what I should substitute so it fits the a U formulas in the back of my book.

Can anyone help?

Integration of:

1/(x*(x-5x^2)^(1/2))

 

[nickman]

The back of the book has a bunch of forumulas with A and U in them, but none match and that x on the bottom is the reason why.

 

[Deleted member 4993]

The question is on integration, a substitution needs to be made before I can use the back of my book to solve it but I'm confused on what I should substitute so it fits the a U formulas in the back of my book.

Can anyone help?

Integration of:

1/(x*(x-5x^2)^(1/2))

Does your problem look like this:

\(\displaystyle \int\frac{1}{x\cdot (x \ - \ 5x^2)} dx\)

If it does - you'll break it up into two fractions using partial fractions first.

 

[nickman]

Yes, it does. Okay, I'll try that and hopefully it works.

 

[nickman]

Does your problem look like this:

\(\displaystyle \int\frac{1}{x\cdot (x \ - \ 5x^2)} dx\)

If it does - you'll break it up into two fractions using partial fractions first.[/quote]


That's the A/ + Bx+C/ kind of stuff, right?

 

[nickman]

I think I get 0 for A, B, C...unless I did something wrong.

 

[galactus]

Let \(\displaystyle x=\frac{1}{u}\) and it whittles down nicely to \(\displaystyle \int\frac{1}{\sqrt{u-5}}du\)

which is easy to integrate.

 

[nickman]

Let \(\displaystyle x=\frac{1}{u}\) and it whittles down nicely to \(\displaystyle \int\frac{1}{\sqrt{u-5}}du\)

which is easy to integrate.

Wait, would that mean let u=1/x ? Making du= -1/x^2 I'm confused as to how it becomes the answer you gave.

 

[Deleted member 4993]

Galactus is right - I missed the ^(1/2) part in the original expression.

 

[nickman]

Galactus is right - I missed the ^(1/2) part in the original expression.

Wait but then

"Wait, would that mean let u=1/x ? Making du= -1/x^2 I'm confused as to how it becomes the answer you gave."

 

[Deleted member 4993]

Galactus is right - I missed the ^(1/2) part in the original expression.

Wait but then

"Wait, would that mean let u=1/x ? Making du= -1/x^2 I'm confused as to how it becomes the answer you gave."

try to do the substitution and show us what you get.

 

[nickman]

[quote="Subhotosh Khan":3cpqvyli]Galactus is right - I missed the ^(1/2) part in the original expression.

Wait but then

"Wait, would that mean let u=1/x ? Making du= -1/x^2 I'm confused as to how it becomes the answer you gave."

try to do the substitution and show us what you get.[/quote:3cpqvyli]

Well after u=1/x, du=-1/x^2, the u=1/x makes it u(x-5x^2)^(1/2) and then....I'm stuck

 

[Deleted member 4993]

Galactus is right - I missed the ^(1/2) part in the original expression.

Wait but then

"Wait, would that mean let u=1/x ? Making du= -1/x^2 I'm confused as to how it becomes the answer you gave."

try to do the substitution and show us what you get.

Well after u=1/x, du=-1/x^2, the u=1/x makes it u(x-5x^2)^(1/2) and then....I'm stuck

dx/[x(x-5x^2)^(1/2)]

= dx/[x * {x^2*(1/x - 5)}[sup:1qkwk0yr]1/2[/sup:1qkwk0yr]]

= 1/[x[sup:1qkwk0yr]2[/sup:1qkwk0yr] * {(1/x - 5)}[sup:1qkwk0yr]1/2[/sup:1qkwk0yr]]

= [1/x[sup:1qkwk0yr]2[/sup:1qkwk0yr] dx]/{(1/x - 5)}[sup:1qkwk0yr]1/2[/sup:1qkwk0yr]

Now can you continue....

Before doing serious work with Calculus, you need to sharpen your skills in Algebra.....

 

[nickman]

Well after u=1/x, du=-1/x^2, the u=1/x makes it u(x-5x^2)^(1/2) and then....I'm stuck[/quote]

dx/[x(x-5x^2)^(1/2)]

= dx/[x * {x^2*(1/x - 5)}[sup:3ty7t0t7]1/2[/sup:3ty7t0t7]]

= 1/[x[sup:3ty7t0t7]2[/sup:3ty7t0t7] * {(1/x - 5)}[sup:3ty7t0t7]1/2[/sup:3ty7t0t7]]

= [1/x[sup:3ty7t0t7]2[/sup:3ty7t0t7] dx]/{(1/x - 5)}[sup:3ty7t0t7]1/2[/sup:3ty7t0t7]

Now can you continue....

Before doing serious work with Calculus, you need to sharpen your skills in Algebra.....[/quote]


Well the whole point of this is to get it into one of the formulas in the back of the book, but once it becomes integral of du/(u-5)^1/2 ...you no longer can do that as some of the forms require like....

integral du/(u*(a^2-u^2))= -1/a ln |(a+ (a^2-u^2)^(1/2)/u | + C

Well anyway, thanks for trying guys.

 

[Deleted member 4993]

Well after u=1/x, du=-1/x^2, the u=1/x makes it u(x-5x^2)^(1/2) and then....I'm stuck

dx/[x(x-5x^2)^(1/2)]

= dx/[x * {x^2*(1/x - 5)}[sup:1dk67qd1]1/2[/sup:1dk67qd1]]

= 1/[x[sup:1dk67qd1]2[/sup:1dk67qd1] * {(1/x - 5)}[sup:1dk67qd1]1/2[/sup:1dk67qd1]]

= [1/x[sup:1dk67qd1]2[/sup:1dk67qd1] dx]/{(1/x - 5)}[sup:1dk67qd1]1/2[/sup:1dk67qd1]

Now can you continue....

Before doing serious work with Calculus, you need to sharpen your skills in Algebra.....[/quote]


Well the whole point of this is to get it into one of the formulas in the back of the book, but once it becomes integral of du/(u-5)^1/2 ...you no longer can do that as some of the forms require like....

integral du/(u*(a^2-u^2))= -1/a ln |(a+ (a^2-u^2)^(1/2)/u | + C

Well anyway, thanks for trying guys.[/quote]

Your book does not have the following formula

\(\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} + C\)

You are kidding .... right....

 

[nickman]

Well after u=1/x, du=-1/x^2, the u=1/x makes it u(x-5x^2)^(1/2) and then....I'm stuck

dx/[x(x-5x^2)^(1/2)]

= dx/[x * {x^2*(1/x - 5)}[sup:3cgoaozg]1/2[/sup:3cgoaozg]]

= 1/[x[sup:3cgoaozg]2[/sup:3cgoaozg] * {(1/x - 5)}[sup:3cgoaozg]1/2[/sup:3cgoaozg]]

= [1/x[sup:3cgoaozg]2[/sup:3cgoaozg] dx]/{(1/x - 5)}[sup:3cgoaozg]1/2[/sup:3cgoaozg]

Now can you continue....

Before doing serious work with Calculus, you need to sharpen your skills in Algebra.....

Well the whole point of this is to get it into one of the formulas in the back of the book, but once it becomes integral of du/(u-5)^1/2 ...you no longer can do that as some of the forms require like....

integral du/(u*(a^2-u^2))= -1/a ln |(a+ (a^2-u^2)^(1/2)/u | + C

Well anyway, thanks for trying guys.[/quote]

Your book does not have the following formula

\(\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} + C\)

You are kidding .... right....[/quote]

Well all the previous forty questions before this use formulas 70+, why would this one shoot back to #1? There has to be a more relevant way to solve this as it pertains to the chapter.

 

[soroban]

Hello, nickman!

\(\displaystyle \displaystyle \int \frac{dx}{x\sqrt{x-5x^2}}\)

Under the radical, complete the square:

. . \(\displaystyle x-5x^2 \;=\;-(5x^2 - x) \;=\;-5\left(x^2 - \tfrac{1}{5}x\right) \;=\;-5\left(x^2 - \tfrac{1}{5}x + \tfrac{1}{100} - \tfrac{1}{100}\right) \;=\;\)

. . . . . . . . \(\displaystyle =\;-5\bigg[\left(x - \tfrac{1}{10}\right)^2 - \tfrac{1}{100}\bigg] \;=\;5\bigg[\tfrac{1}{100} - \left(x - \tfrac{1}{10}\right)^2\bigg]\)

. . \(\displaystyle \text{Then: }\;\sqrt{x-5x^2} \;=\;\sqrt{5}\,\sqrt{\left(\tfrac{1}{10}\right)^2 - \left(x - \tfrac{1}{10}\right)^2 }\)


\(\displaystyle \displaystyle \text{The integral becomes: }\;\frac{1}{\sqrt{5}}\int \frac{dx}{x\sqrt{(\frac{1}{10})^2 - (x-\frac{1}{10})^2}}\)


\(\displaystyle \text{Let: }\:x-\tfrac{1}{10} \;=\;\tfrac{1}{10}\sin\theta \quad\Rightarrow\quad x \;=\;\tfrac{1}{10}(\sin\theta + 1) \quad\Rightarrow\quad dx \:=\:\tfrac{1}{10}\cos\theta\,d\theta\)


\(\displaystyle \displaystyle \text{Substitute: }\;\frac{1}{\sqrt{5}} \int \frac{\frac{1}{10}\cos\theta\,d\theta}{\frac{1}{10}(\sin\theta + 1)\cdot \frac{1}{10}\cos\theta} \;\;=\;\;\frac{10}{\sqrt{5}} \int \frac{d\theta}{\sin\theta + 1} \;\;=\;\;2\sqrt{5}\int\frac{d\theta}{1 + \sin\theta}\)


\(\displaystyle \text{Multiply by }\frac{1-\sin\theta}{1-\sin\theta}\!:\quad2\sqrt{5}\int\frac{d\theta}{1 + \sin\theta}\cdot\frac{1-\sin\theta}{1-\sin\theta} \;\;=\;\;2\sqrt{5}\int \frac{1-\sin\theta}{1-\sin^2\!\theta}\,d\theta\)


. . . . . . . . . . . . . . . \(\displaystyle \displaystyle =\;2\sqrt{5}\int \frac{1-\sin\theta}{\cos^2\!\theta}\,d\theta \;\;=\;\; 2\sqrt{5}\int\left(\frac{1}{\cos^2\!\theta} - \frac{\sin\theta}{\cos^2\!\theta}\right)d\theta\)


. . . . . . . . . . . . . . . \(\displaystyle =\; 2\sqrt{5} \int\left(\sec^2\!\theta - \sec\theta\tan\theta)\,d\theta\)

I'll let you finish it . . .

 

[BigGlenntheHeavy]

\(\displaystyle \int\frac{dx}{x(x-5x^2)^{1/2}} \ = \ \int\frac{dx}{x[x(1-5x)]^{1/2}} \ = \ \int\frac{dx}{x^{3/2}(1-5x)^{1/2}}\)

\(\displaystyle Now. \ let \ u \ = \ (1-5x)^{1/2}, \ du \ = \ (1/2)(1-5x)^{-1/2}(-5)dx, \ \implies \ \frac{2udu}{-5} \ = \ dx\)

\(\displaystyle Also, \ u \ = \ (1-5x)^{1/2}, \ \implies \ u^2 \ = \ 1-5x, \ \implies \ x \ = \ \frac{1-u^2}{5}, \ x^{3/2} \ = \ \frac{(1-u^2)^{3/2}}{5^{3/2}}\)

\(\displaystyle Now, \ putting \ that \ all \ together, \ we \ get: \ -2\sqrt5\int\frac{du}{(1-u^2)^{3/2}}\)

\(\displaystyle Next \ sub: \ Let \ u \ = \ sin(\theta), \ du \ = \ cos(\theta)d\theta\)

\(\displaystyle This \ gives: \ -2\sqrt5\int\frac{cos(\theta)d\theta}{[1-sin^2(\theta)]^{3/2}} \ = \ -2\sqrt5\int\frac{cos(\theta)d\theta}{cos^3(\theta)} \ = \ -2\sqrt5\int\frac{d\theta}{cos^2(\theta)}\)

\(\displaystyle =-2\sqrt5\int sec^2(\theta)d\theta \ = \ -2\sqrt5 tan(\theta) \ +C\)

\(\displaystyle Now. \ resubbing, \ we \ get: \ -2\sqrt5 \frac{u}{\sqrt(1-u^2)}+C \ = \ -2\sqrt5 \frac{(1-5x)^{1/2}}{\sqrt{1-(1-5x)}}+C\)

\(\displaystyle = \ -2\frac{(1-5x)^{1/2}}{x^{1/2}}+C, \ QED, \ I'll \ leave \ the \ check \ up \ to \ whomever.\)

 

[galactus]

The sub \(\displaystyle x=\frac{1}{u}, \;\ dx=\frac{-1}{u^{2}}du\)

\(\displaystyle \int\frac{1}{\frac{1}{u}\cdot \sqrt{\frac{1}{u}-\frac{5}{u^{2}}}}\cdot \frac{-1}{u^{2}}du\)

The algebra hammers it into \(\displaystyle \int\frac{1}{\sqrt{u-5}}du\)

Now, integrate and resub.

The worst thing for many calc students is the algebra. Try working at this one and see if you can whittle it down as I showed.