# Integration problem

## Homework Statement

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

## 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

help~

vela
Staff Emeritus
Homework Helper

## Homework Statement

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

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

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
Homework Helper
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
Homework Helper
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
Homework Helper
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
Gold Member
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
Homework Helper
Char.Limit said:
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.

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

try the substitution

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

Cyosis
Homework Helper
mk200789 said:
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.

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

Cyosis
Homework Helper
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
Staff Emeritus