MHB Integral solve by using trig. substitution

[karush]

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?

 

[Ackbach]

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]

you mean this?? $81\int u^3 du$

 

[alyafey22]

You can solve it without tri-substitution using $$t=9-x^2$$

 

[Ackbach]

you mean this?? $81\int u^3 du$

Yep, that's it.

 

[karush]

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]

We are given:

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

and told to use the substitution:

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

And so the integral becomes:

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

So we obtain:

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

Using a Pythagorean identity, we may then write:

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

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

 

[Ackbach]

We are given:

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

and told to use the substitution:

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

And so the integral becomes:

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

So we obtain:

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

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

 

[karush]

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

$\cos\theta=\frac{\sqrt{9-x^2}}{3}$

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

 

[karush]

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]

$\cos\theta=\frac{\sqrt{9-x^2}}{3}$

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]

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

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
$