Integrate (x^2 + y^2)^(3/2) dy = 2[(x^2 + y^2)^(5/2)] / 10y - Expert Help & Tips

[teng125]

integra of (x^2 + y^2)^(3/2) dy is it equals to 2[(x^2 + y^2)^(5/2)] / 10y
??

pls let me know
thanx

 

[Hootenanny]

integra of (x^2 + y^2)^(3/2) dy is it equals to 2[(x^2 + y^2)^(5/2)] / 10y
??

pls let me know
thanx

Nope, that is not correct. Perhaps if you showed your steps...

 

[teng125]

i just integrate as (x^2 + y^2)^(5/2) / (5/2)(2y) .may i know how to do??

 

[Gagle The Terrible]

How to do what ?

 

[teng125]

pls show me how to integrate the eqn above

 

[Gagle The Terrible]

try to substitute u = x^2 + y^2

Substitutions are a handy trick when the denominator is the derivative of the numerator.

 

[Hootenanny]

HINT: You may have to substitute more than once.

 

[teng125]

pls show,how to substitute more than once

thanx

 

[Gagle The Terrible]

Perseverance, my friend, is the key to math. enlightment. In the worst case scenario, erase and start over. Try to figure it out by yourself, and if you really can't get to the answer, show what went wrong and then maybe...

 

[teng125]

i subs u=x^2 + y^2 and differentiate then get du=2x dx
then integra sqrt[u/(u-y^2)] du and fromhere onwards i don't know how to integra wif respect to u

pls help

thanx

 

[Gagle The Terrible]

This is where the wisdom of hootnanny comes in play. Once you find du , you must find an expression with du where dy/5y can be "eliminated". You know that du = 2ydy. Therefore du/(10y^2) = dy/5y .

From that, substitue y^2 again ...

EDIT : Yo uare integrating with respect to what variable ??
Plz specify ...I don't think I got your question right when I saw your last post.

 

[Gagle The Terrible]

If you are trying to integrate wrt x, try trig substitutions. If you are trying to integrate wrt y, try the substitution i mentionned earlier .

 

[Hootenanny]

If you are trying to solve;

\int (x^{2} + y^{2})^{\frac{3}{2}} \; dy

The after your double substitution you should end up with something like this (unless I have made a mistake somewhere);

\int \frac{U^{\frac{3}{2}}}{2(U - x^{2})^{\frac{1}{2}}} \; du

 

[teng125]

ya,i got stuck here also,how to proceed??

 

[VietDao29]

Let a be some constant.
In an integral, when you meet the form of a² + x², then you may use the trig substitution x = a \tan \theta
If you meet the form of a² - x², then you can to use a trig substitution x = a \sin \theta
If you meet the form of x² - a², then you can to use a trig substitution x = a \sec \theta.
You can take a look here.
So in this problem, what you should do is to use the trig substitution y = x \tan \theta.
\Rightarrow dy = \frac{x d ( \theta )}{\cos ^ 2 \theta}
Ok, let's see if you can go from here. :)
Hint: You should use the identity 1 + tan²x = sec²x.
By the way, I advise you to re-read the book. Read again from the very first lesson of integral, if possible, and make sure you understand everything. It won't do you any harm, I promise. Just for your own sake. :)

 

[teng125]

thanx...

 

[arildno]

An even simpler choice would be to set y=x\sinh(u)[/tex]

 

[teng125]

may i know how to integrate sec^5(tita) d(tita) pls

 

[teng125]

i got x^4 cosh^4(u) du but how can i integra from here??
what shoulc i do with cosh^4(u) du??

pls help

 

[VietDao29]

Integrating odd power of secant, and cosecant is a bit tricky. You can use Integrate by Parts to solve this. Say you want to integrate:
I_n = \int \sec ^ n \theta d( \theta ).
Let u = \sec ^ {n - 2} \theta \quad \mbox{and} \quad dv = \sec ^ 2 \theta d ( \theta )
\Rightarrow du = (n - 2) \sec ^ {n - 2} n \theta \tan \theta d( \theta ) \quad \mbox{and} \quad v = \tan \theta
So, we have:
I_n = \int \sec ^ n \theta d( \theta ) = \sec ^ {n - 2} \theta \tan \theta - (n - 2) \int \sec ^ {n - 2} \tan ^ 2 \theta d( \theta ) = \sec ^ {n - 2} \theta \tan \theta - (n - 2) \int \sec ^ {n - 2} \left( \sec ^ 2 - 1 \right) d( \theta )
= \sec ^ {n - 2} \theta \tan \theta - (n - 2) \int \sec ^ n \theta d( \theta ) + (n - 2) \int \sec ^ {n - 2} \theta d( \theta ) = \sec ^ {n - 2} \theta \tan \theta - (n - 2) I_n + (n - 2) I_{n - 2}
Rearranging it gives:
(n - 1) I_n = \sec ^ {n - 2} \theta \tan \theta + (n - 2) I_{n - 2} or:
I_n = \frac{1}{n - 1} \left( \sec ^ {n - 2} \theta \tan \theta + (n - 2) I_{n - 2} \right).
Appling it here, we have:
I_5 = \frac{1}{4} \left( \sec ^ {3} \theta \tan \theta + (3) I_{3} \right).
Can you go from here?
Hint: You may need to apply it once more, and use the fact that:
\int \sec \theta d( \theta ) = \ln | \sec \theta + \tan \theta | + C
--------------------
You can also use the u substitution:
u = \sin \theta to integrate \int \sec ^ 5 \theta d( \theta ) = \int \frac{1}{\cos ^ 5 \theta} d( \theta ), since the power of cos is odd. And then use Partial fraction. Something like this:
\int \sec ^ 5 \theta d( \theta ) = \int \frac{1}{\cos ^ 5 \theta} d( \theta ) = \frac{\cos \theta}{\cos ^ 6 \theta} d( \theta ) = \frac{1}{(1 - u ^ 2) ^ 3} du = ...
Can you go from here? :)

 

[teng125]

i prefern the use substitution but i cannot go from there

 

[VietDao29]

Uhmm, You only stuck at the Partial Fraction part right? You can take a look at this page. Read it thoroughly before trying to tackle this problem.
Try to solve for the constants A, B, C, D, E, F in the equation:
\frac{1}{(1 - u ^ 2) ^ 3} = \frac{1}{(1 - u) ^ 3 (1 + u) ^ 3} = \frac{A}{1 - u} + \frac{B}{(1 - u) ^ 2} + \frac{C}{(1 - u) ^ 3} + \frac{D}{1 + u} + \frac{E}{(1 + u) ^ 2} + \frac{F}{(1 + u) ^ 3}.
I'd prefer the first way though. :)