Integration problem

[mk200789]

Homework Statement

integrate (x^3)sqrt(x^2 + 8) dx

Homework Equations

The Attempt at a Solution

let x = 2sqrt(2)tan(t) ==> dx= 2sqrt(2)sec^2(t) dt

=int (x^3)sqrt(x^2 + 8) dx
=int (16sqrt(2)tan^3(t)) sqrt(8tan^2(t) + 8) (2sqrt(2)sec^2(t)) dt
=int (32 sqrt(2)) (tan^3(t)) sec^3(t) dt

my problem is till this point i compared it with the the prof. working
and he had this instead : int (16sqrt(2)tan^3(t))(2sqrt(2)sec(t)) dt

help~

 

[vela]

Homework Statement

integrate (x^3)sqrt(x^2 + 8) dx

Homework Equations

The Attempt at a Solution

let x = 2sqrt(2)tan(t) ==> dx= 2sqrt(2)sec^2(t) dt

=int (x^3)sqrt(x^2 + 8) dx
=int (16sqrt(2)tan^3(t)) sqrt(8tan^2(t) + 8) (2sqrt(2)sec^2(t)) dt
=int (32 sqrt(2)) (tan^3(t)) sec^3(t) dt

my problem is till this point i compared it with the the prof. working
and he had this instead : int (16sqrt(2)tan^3(t))(2sqrt(2)sec(t)) dt

Your work is fine. It looks like your professor didn't substitute for dt.

 

[julian92]

I didn't really take a look at your solution

But in order to solve the integral ,, just substitute >>> u = 8 + x^2

it's straight forward ;)

 

[Cyosis]

I didn't really take a look at your solution

But in order to solve the integral ,, just substitute >>> u = 8 + x^2

it's straight forward ;)

It may be straightforward but this substitution is pretty useless.

Integration by parts is the way to go here and you only have to do it once.

 

[Gib Z]

It may be straightforward but this substitution is pretty useless.

Integration by parts is the way to go here and you only have to do it once.

The solution comes out almost straight away with the substitution actually.

 

[Cyosis]

The solution comes out almost straight away with the substitution actually.

While useless may have been a bit strong I still feel you don't gain much by doing that substitution. You will have to do integration by parts after wards, which you may as well do right away.

 

[Char. Limit]

The solution comes out almost straight away with the substitution actually.

How exactly? du does not equal x^3, it equals 2x. Rather useless.

 

[Cyosis]

How exactly? du does not equal x^3, it equals 2x. Rather useless.

The idea is that x^2=u-8 and we can write x^3 as x^2*x. You can then combine them. However in my eyes that just sends you back to the original problem. We may be going a bit off topic here.

 

[mk200789]

SO...my working so far is correct? or the prof's correct?:|

 

[Dickfore]

try the substitution

$$t = \sqrt{x^{2} + 8}$$

 

[Cyosis]

SO...my working so far is correct? or the prof's correct?:|

You are correct. However unless the exercise asks you to specifically use a trigonometric substitution you should really use integration by parts.

 

[mk200789]

thanks. yeh i was told to use trig sub:)

 

[Cyosis]

I just noticed that you made a minor mistake regarding the constant factors.

=int (16sqrt(2)tan^3(t)) sqrt(8tan^2(t) + 8) (2sqrt(2)sec^2(t)) dt

This is correct.

=int (32 sqrt(2)) (tan^3(t)) sec^3(t) dt

Here you made a mistake when it comes to multiplying the constant roots.

 

[vela]

While useless may have been a bit strong I still feel you don't gain much by doing that substitution. You will have to do integration by parts after wards, which you may as well do right away.

No integration by parts is needed with the substitution. You just multiply it out to get two terms of the form u^n.