Determine the integral `int x^3*sqrt(x^2 - 25) dx`

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justaguide eNotes educator| Certified Educator

The integral `int x^3*sqrt(x^2 - 25) dx` has to be determined.

`int x^3*sqrt(x^2 - 25) dx`

= `int x^2*x*sqrt(x^2 - 25) dx`

Let `x^2 - 25 = y`

`x^2 = y + 25`

`dy/dx = 2x`

`dy/2 = x*dx`

The integral `int x^3*sqrt(x^2 - 25) dx` can be written as:

`int (y + 25)*sqrt y*(1/2)*dy`

= `(1/2)*int y*sqrt y + 25*sqrt y dy`

= `(1/2)*int y^(3/2) dy + (25/2)*int y^(1/2) dy`

= `(1/2)*(y^(5/2))/(5/2) + (25/2)*y^(3/2)/(3/2) + C`

= `(1/5)*(y^(5/2)) + (25/3)*y^(3/2) + C`

Substitute `y = x^2 - 25`

= `(1/5)*(x^2 - 25)^(5/2) + (25/3)*(x^2 - 25)^(3/2) + C`

The integral `int x^3*sqrt(x^2 - 25) dx` is `(1/5)*(x^2 - 25)^(5/2) + (25/3)*(x^2 - 25)^(3/2) + C`

anhtranru | Student

Hey Moo, this will be a pretty long and rather complicated answer, but I will try my best to present it to you. The technique that I used for this integral is called trig substitution. I hope you already learned it. 

So firstly, the part (x^2-25) under the square root remind me of the relationship between sec and tan. We know that (tanx)^2+1=(secx)^2, with manipulation, we have (tanx)^x= (secx)^2-1. 

Therefore, my trig substitution is x=5sec(u)



Now plug this into our expression, we have

integral ((5sec)^3*sqrt(25(sec(u))^2-25))*5tan(u)sec(u)du

you notice that the we can factor out the 25 in the part under the square root. So our integral will look like this. Also, we can pull the constant outside the integral, to make the integral look neater. 

5^5 (integral ((sec u)^3*sqrt((sec u)^2-1))tan(u)sec(u)du

And we did mention before that tanx=sqrt((sec u)^2-1)

so the integral now is this 

5^5 (int. (sec^3(u) *tan(u)*tan(u)*sec(u)du)

5^5 (int.(sec^4(u)*tan^2(u)du)

Now we have to do another substitution. (I know it's a long solution, but bear with me ^^)

let w=tan(u)


so the integral is now become this

5^5 int. (sec^2(u)*(w^2))dw

we know that sec^2(u)=(1+w) derived from the relationship of tan and sec.

=> 5^5 int. ((1+w)*w^2)dw

Now just distributed.

=> 5^5 int. (w^2+w^3)dw

Now this integral is easy, just exponent rule

=> 5^5*((w^3)/3 + (w^4)/4) +C

But now, we have to 'convert' everything back to x, because our original integral was given with respect to x. 

so first let convert it back to u 

=>5^5*((tan^3(u))/3 +(tan^4(u))/4) +C

to convert u back to x, we need to construct the trig triangle using the fact we set x=sec(u) before.

with some manipulation, tan(u)=(sqrt(x^2-25))/5

so the final answer should be 

5^5*(((sqrt(x^2-25))/(5*3))^3+((sqrt(x^2-25))/(5*4))^4) +C

That's it! I hope you will able to read the way I represent the answer. Good luck!

moocow554 | Student

`intx^3sqrt(x^2-25)` is the question.... sorry!!!