[SOLVED]Integral solve by using trig. substitution

karush

Well-known member
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

Indicium Physicus
Staff member
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
 you mean this?? $81\int u^3 du$

ZaidAlyafey

Well-known member
MHB Math Helper
You can solve it without tri-substitution using $$\displaystyle t=9-x^2$$

Staff member

karush

Well-known member
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

Staff member
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
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
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
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

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

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

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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