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[SOLVED] Integral solve by using trig. substitution

karush

Well-known member
Jan 31, 2012
2,714
Evaluate integral by using $x=3\sin{\theta}$
$\int{x^3\sqrt{9-x^2}}\ dx$

substituting

$\int{27\sin^3{\theta}}\sqrt{9-9\sin^2{\theta}}\Rightarrow
81\int\sin^3\theta\cos\theta\ dx$

since the power of sine is odd then

$81\int\sin^2{\theta}\cos{\theta}\sin{\theta}\ dx$

$81\int\left(1-\cos^2{\theta}\right)\cos{\theta}\sin{\theta}\ dx$

hope ok so far but next steps????:confused:
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Re: Integral solve by using trig substiturion

Evaluate integral by using $x=3\sin{\theta}$
$\int{x^3\sqrt{9-x^2}}\ dx$

substituting

$\int{27\sin^3{\theta}}\sqrt{9-9\sin^2{\theta}}\Rightarrow
81\int\sin^3\theta\cos\theta\ dx$
At this point, why not just do $u=\sin(\theta)$?

[EDIT] See MarkFL's post for a correction.
 

karush

Well-known member
Jan 31, 2012
2,714
you mean this??
$81\int u^3 du$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You can solve it without tri-substitution using \(\displaystyle t=9-x^2\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191

karush

Well-known member
Jan 31, 2012
2,714
ok but the exercise wants us to: "Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle" W|F steps completely left me in dense fog...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given:

\(\displaystyle \int x^3\sqrt{9-x^2}\,dx\)

and told to use the substitution:

\(\displaystyle x=3\sin(\theta)\,\therefore\, dx=3\cos(\theta)\,d\theta\)

And so the integral becomes:

\(\displaystyle \int \left(3\sin(\theta) \right)^3\sqrt{9-\left(3\sin(\theta) \right)^2}3\cos(\theta)\,d\theta\)

So we obtain:

\(\displaystyle 3^5\int\sin^3(\theta)\cos^2(\theta)\,d \theta\)

Using a Pythagorean identity, we may then write:

\(\displaystyle 3^5\int\left(\cos^2(\theta)-\cos^4(\theta) \right)\sin(\theta)\,d\theta\)

Using a right triangle where \(\displaystyle \sin(\theta)=\frac{x}{3}\), what then is \(\displaystyle \cos(\theta)\)?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
We are given:

\(\displaystyle \int x^3\sqrt{9-x^2}\,dx\)

and told to use the substitution:

\(\displaystyle x=3\sin(\theta)\,\therefore\, dx=3\cos(\theta)\,d\theta\)

And so the integral becomes:

\(\displaystyle \int \left(3\sin(\theta) \right)^3\sqrt{9-\left(3\sin(\theta) \right)^2}3\cos(\theta)\,d\theta\)

So we obtain:

\(\displaystyle 3^5\int\sin^3(\theta)\cos^2(\theta)\,d \theta\)
Ah, very good. Didn't catch that error in the OP. It's $\cos^{2}(\theta)$, not $\cos(\theta)$.
 

karush

Well-known member
Jan 31, 2012
2,714
Using a right triangle where \(\displaystyle \sin(\theta)=\frac{x}{3}\), what then is \(\displaystyle \cos(\theta)\)?
$\cos\theta=\frac{\sqrt{9-x^2}}{3}$

I think anyway. but don't we plug this later?
 

karush

Well-known member
Jan 31, 2012
2,714
I will return to this later have to go now... but all these is very helpful... to tough to do by myself..
....to be continued...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\cos\theta=\frac{\sqrt{9-x^2}}{3}$

I think anyway. but don't we plug this later?
Yes, I just saw you asking about it when I was alerted that reponses had been post while I was composing my reply. You will want to use that after you integrate, to get the anti-derivative in terms of $x$.
 

karush

Well-known member
Jan 31, 2012
2,714
$\displaystyle
3^5\int\left(\cos^2(\theta)-\cos^4(\theta) \right)\sin(\theta)\,d\theta$

well from this if $u=\cos{\theta}$ and $du = -\sin{\theta}$ and $\cos{\theta} = \frac{\sqrt{9-x^2}}{3}$ then

$\displaystyle
-3^5\int\left(u^2-u^4\right)\sin(\theta)\,d\theta
\Rightarrow
-3^5\left[\frac{u^3}{3}-\frac{u^5}{5}\right]+C
\Rightarrow
\frac{-(x^2+6)\cdot(9-x^2)^{3/2}}{5}+C
$
 
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