1. ## Trigonometric Subsitution

Hello peeps,

What I need help on is clarifying if the answer for this integral is correct because I have did it in two ways and I'm not sure if both of them are correct.

$\int x^3\sqrt{9-x^2}\,dx$

$x = 3sin\Theta$
$dx = 9sin^2\Theta d\Theta\$

for the square root I have:
$\sqrt{9 - x^2}\ = \sqrt{9- 9 sin^2\Theta}\ = \sqrt{9(1-sin^2\Theta)}\ = \sqrt{9cos^2\Theta}\ = 3cos\Theta$

$
\int 27sin^3\Theta 3cos\Theta 9sin^2\Theta d\Theta\
=
\int 729sin^3\Theta cos\Theta sin^2\Theta d\Theta\
$

From here I did the equation 2 ways, please let me know if one of the ways is incorrect.

1st way:
$
\int 27sin^3\Theta 3cos\Theta 9sin^2\Theta d\Theta\
$

$
u = sin\Theta
du = cos\Theta d\Theta
$

$
729\int u^3 u^2 du\
=
729\int u^5 du\
= 729 \frac{u^6}6 +C
= 729 \frac{sin^6}6 +C
= 729 \frac{\frac{x}{3}^6}{6} + C
$

2nd way

$
\int 729sin^3\Theta 3cos\Theta sin^2\Theta d\Theta\
=
\int 729sin^4\Theta cos\Theta sin\Theta d\Theta\
$

$
u = cos\Theta
du = -sin\Theta d\Theta
$

$
\int 729 (1-cos^2\Theta)^2 cos\Theta sin\Theta d\Theta\
=
-\int 729 u(1-u^2)^2 du
$

$
= -(1 - 2u^2 + u^4)u
= (1 + 2u^2 - u^4)u
= u + 2u^3 - u^5
$

$
= \frac{u^2}{2} + 2\frac{u^4}{4} - \frac{u^6}{6}
= \frac{cos\Theta^2}{2} + 2\frac{cos\Theta^4}{4} - \frac{cos\Theta^6}{6} + C
$

$
=729 \frac{\frac{\sqrt{9-x^2}^2}{3}}{2} + \frac{\frac{\sqrt{9-x^2}^4}{3}}{4} +
\frac{\frac{\sqrt{9-x^2}^6}{3}}{6} + C
$

2. Originally Posted by ff4930
Hello peeps,

What I need help on is clarifying if the answer for this integral is correct because I have did it in two ways and I'm not sure if both of them are correct.

$\int x^3\sqrt{9-x^2}\,dx$

$x = 3sin\Theta$
$\color {red}dx = 9sin^2\Theta d\Theta\$ ...
you messed up in this line, so everything afterwards is incorrect

what is the derivative of $3 \sin \theta$ ?

3. .... $3 cos \Theta$
I feel like an idiot.
I was thinking of x^2 for some reason, thanks anyways.

4. Originally Posted by ff4930
.... $3 cos \Theta$
I feel like an idiot.
we all make mistakes. try again

5. Ok, on the second try...

I get

$
\int 3^3 sin^3\Theta 3cos\Theta 3cos\Theta d\Theta\
=
\int 3^5 sin^2\Theta sin\Theta cos^2\Theta d\Theta\
$

$
=
3^5 \int (1-cos^2\Theta) sin\Theta cos^2\Theta d\Theta\
let u = cos\Theta
du = -sin\Theta d\Theta
$

$
-3^5 \int (1-u^2)u^2du\
=
3^5 \int u^2+u^4du\
$

$
= 3^5 \int \frac{u^3}{3}+\frac{u^5}{5}du\
= 3^5 \frac{cos\Theta^3}{3}+\frac{cos\Theta^5}{5}du
$

$
= 3^5 \frac{\frac{\sqrt{9-x^2}^3}{3}}{3} +
\frac{\frac{\sqrt{9-x^2}^5}{3}}{5}
$

Is this correct? is an even problem on my textbook so I cant look it up, thanks for your patience.

6. Originally Posted by ff4930
Ok, on the second try...

I get

$
\int 3^3 sin^3\Theta 3cos\Theta 3cos\Theta d\Theta\
=
\int 3^5 sin^2\Theta sin\Theta cos^2\Theta d\Theta\
$

$
=
3^5 \int (1-cos^2\Theta) sin\Theta cos^2\Theta d\Theta\
let u = cos\Theta
du = -sin\Theta d\Theta
$

$
-3^5 \int (1-u^2)u^2du\
=
3^5 \int u^2+u^4du\
$

$
= 3^5 \int \frac{u^3}{3}+\frac{u^5}{5}du\
= 3^5 \frac{cos\Theta^3}{3}+\frac{cos\Theta^5}{5}du
$

$
= 3^5 \frac{\frac{\sqrt{9-x^2}^3}{3}}{3} +
\frac{\frac{\sqrt{9-x^2}^5}{3}}{5}
$

Is this correct? is an even problem on my textbook so I cant look it up, thanks for your patience.
you can actually check your answer here. just type in your function, you would enter "(x^3)*Sqrt[9 - x^2]" --without the quotes of course. i will look over your work in a little while

7. Originally Posted by ff4930
$\int x^3\sqrt{9-x^2}\,dx$
Are you sure you need a trig. substitution?

They're nice, but in this case, I would do the following

According to the change of variables defined by $u=9-x^2\implies du=-2x\,dx$, the integral becomes to

$\int x^3\sqrt{9-x^2}\,dx=-\frac12\int(9-u)\sqrt u\,du$

Does that make sense?

P.S.: by the way ff4930, you can use \theta, looks better.

8. Originally Posted by Krizalid
Are you sure you need a trig. substitution?

They're nice, but in this case, I would do the following

According to the change of variables defined by $u=9-x^2\implies du=-2x\,dx$, the integral becomes to

$\int x^3\sqrt{9-x^2}\,dx=-\frac12\int(9-u)\sqrt u\,du$

Does that make sense?

P.S.: by the way ff4930, you can use \theta, looks better.
that's the way i would have done it. i don't like trig. substitution that much. but ff4930 wants it. i think he's required to do it that way.

Originally Posted by ff4930
Ok, on the second try...

I get

$
\int 3^3 sin^3\Theta 3cos\Theta 3cos\Theta d\Theta\
=
\int 3^5 sin^2\Theta sin\Theta cos^2\Theta d\Theta\
$

$
=
3^5 \int (1-cos^2\Theta) sin\Theta cos^2\Theta d\Theta\
let u = cos\Theta
du = -sin\Theta d\Theta
$

$
-3^5 \int (1-u^2)u^2du\
=
\color {red} 3^5 \int u^2+u^4du\
$
...
you messed up the signs in the last line shown here

9. How you get (9-u) from x^3?

Also, the problems are from a textbook and is on the section of trig. substitution.

First time using the math tags, took me half hour to do the first post =[

rats, didn't see the sign error, other than that the steps were fine?

10. Originally Posted by ff4930
$\int x^3\sqrt{9-x^2}\,dx$
Well... the integrator gives $-\frac15(x^2+6)(9-x^2)^{3/2}+k$

So, you should get that

Sometimes, trigs. substitutions are really cool, but it depends of the integral, of course.

11. Originally Posted by ff4930
How you get (9-u) from x^3?

Also, the problems are from a textbook and is on the section of trig. substitution.

First time using the math tags, took me half hour to do the first post =[

rats, didn't see the sign error, other than that the steps were fine?
he said $u = 9 - x^2 \implies x^2 = 9 - u$. since $du = -2x ~dx$, the du took away one of the x's from the x^3 and we were left with x^2. now we just replaced the x^2 with the function of u i mentioned above

12. Originally Posted by Krizalid
Well... the integrator gives $-\frac15(x^2+6)(9-x^2)^{3/2}+k$

So, you should get that
not necessarily. sometimes the result of the same integral can have hugely different forms depending on what integration technique you use. often you can simplify one form to get the other, but sometimes you can't, the difference is absorbed in the arbitrary constant.

we just need to take it step by step here and make sure ff4930 made no errors. which i am pointing out has he goes along...

Sometimes, trigs. substitutions are really cool, but it depends of the integral, of course.
i agree

13. Thanks guys, I really appreciate your help.

I've had this math professor since calculus 1 and he has always let us student leave the answer in such a mess and does not require us to simplify the answer. So now I'm so used to just to leaving a really long answer and not bother simplifying.

14. Originally Posted by Jhevon
not necessarily. sometimes the result of the same integral can have hugely different forms depending on what integration technique you use. often you can simplify one form to get the other, but sometimes you can't, the difference is absorbed in the arbitrary constant.
Of course dear Jhevon, I knew that

It's true, sometimes the integrator gives awful answers, and we get beauty ones