# Integration Using Trigonometric Substitution Help Needed!

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1. Feb 22, 2015

### Airp

1. The problem statement, all variables and given/known data
Integral of $$x^3\sqrt{x^2+16}dx$$
$$1/5(x^2+16)^{5/2} -16/3(x^2+16)^{1/2}+C$$
2. Relevant equations
x=atanθ

3. The attempt at a solution

Mod note: The integral is $\int x^3 \sqrt{x^2 + 16} dx$
The published answer is $1/5(x^2+16)^{5/2} -16/3(x^2+16)^{1/2}+C$

To the OP: Using  (i.e., with a space between the $symbols) causes the browser to not render what's between the$ pairs. For inline LaTeX, you can use # # (no space) at the beginning and end.

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Last edited: Feb 22, 2015
2. Feb 22, 2015

### Faris Shajahan

Please explain what you are trying to do

3. Feb 22, 2015

### Airp

I'm trying to find the integral of the equation, but it doesn't give the same answer as the one I'm supposed to get, and I don't know what I'm doing wrong. I'm using trigonometric substitution which lets x=4sinΘ.

4. Feb 22, 2015

### Faris Shajahan

In the second step after you did the substitution, the whole things becomes,
4*((tan0^2)+1)^(1/2)=4sec0 and not 4sec0^2

5. Feb 22, 2015

### Faris Shajahan

can you post the question? Like the integral you are trying to find the answer for?

6. Feb 22, 2015

### Airp

No problem! I tried using Latex, but it doesn't work...It's #6

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7. Feb 22, 2015

### Faris Shajahan

I might be able to help you....I love integration...but I don't understand anything from the picture you posted

8. Feb 22, 2015

### Staff: Mentor

See my previous post. It's fairly clear which integral the OP is trying to find.

9. Feb 22, 2015

### Staff: Mentor

See my edit to post #1.

10. Feb 22, 2015

### Airp

Thank You Mark! I'm still learning to use Latex... I changed it and it should appear ok, now!

11. Feb 22, 2015

### Faris Shajahan

Got it!!!

12. Feb 22, 2015

### Airp

Ok, I'm currently trying, correcting the mistake I made with my sec^2. It looks good!

13. Feb 22, 2015

### Airp

I can officially say that it works now! Thank you Faris! Such a "minor" mistake that really changed everything in the problem!

14. Feb 22, 2015

### TheRedDevil18

He meant Question 6 from the picture

15. Feb 22, 2015

### TheRedDevil18

Your trig substitution is correct. When you square root, you should get 4sec(x) (I'll use x instead of theta for simplicity)
That will leave you with,
64tan3(x) * 4sec(x)

or simply
256tan3(x)*sec(x)

Break the tan3(x) up, you get
256tan2(x)*sec(x)*tan(x)

I wont do the rest for you but I'll give you a clue. It involves a trig identity and a u substitution

16. Feb 22, 2015

### Airp

Thank you! It works now, my mistake was simply that I forgot to take the sqrt of sec^2

17. Feb 22, 2015

### Faris Shajahan

hey, if trigonometric substitution is not necessary, it is much easier to put x^2+16=t.......=>x^2=t-16.......dx=dt/2x.......x^3=x*x2.....
then integral is $\int { x*(t-16)*\sqrt(t)*dt/2x}$ => $\int {(t-16)\sqrt(t) dt/2}$ ............now expand normally...