trig. substitution

**saiyanmx89** · March 1st 2009, 07:31 AM

$\int x \sqrt{16-4x^2}dx => \int x \sqrt{4^2-(2x)^2}dx$

$2x = 4sin\theta$

$2dx = 4cos\theta d\theta$

$dx = 2cos\theta d\theta$

$= \int x \sqrt{4^2-(4sin\theta)^2}*2cos\theta d\theta$

$= 2\int x \sqrt{16(1-sin^2\theta)}*cos\theta d\theta$

$= 2 \int x(4cos\theta cos\theta)d\theta$

$= 2 \int x(4cos^2\theta)d\theta$

According to my substitution, should I let x=2sin $\theta$ and substitute that back in here??

$= 2 \int 2sin\theta(4cos^2d\theta)$ ???

If so, where should I go from there?

**Nokio720** · March 1st 2009, 09:37 AM

$<br /> <br /> = 16 \int sin\theta(cos^2\theta d\theta)<br />$

use u substitution and let $u=cos \theta$

after integration use your initial supposition that $2x = 4sin\theta$ to rewrite your answer in terms of x.

**Thomas154321** · March 1st 2009, 09:42 AM

$2 \int 2sin\theta(4cos^2\theta)d\theta = 16 \int sin\theta(cos^2\theta)d\theta$

$= 16 \left[ {\frac{cos^3\theta}{3}} \right] +c$

(Check it works by differentiating back.) In general:

$\int f'(x)f^n(x) dx = \frac{f^{n+1}(x)}{n+1} +c$

**saiyanmx89** · March 1st 2009, 09:50 AM

how do I rewrite:
$\frac{-16cos^3\theta}{3} +C$
in terms of x? I don't understand.

**Soroban** · March 1st 2009, 10:23 AM

Hello, saiyanmx89!

How do I rewrite: $-\frac{16}{3}\cos^3\!\theta +C$ in terms of $x$ ?

Is this your first Trig Substitution?

We had: . $2x \:=\:4\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{hyp}$

Using Pythagorus, we find that: . $adj \:=\:\sqrt{4-x^2}$

. . Hence: . $\cos\theta \:=\:\frac{\sqrt{4-x^2}}{2}$

Substitute: . $-\frac{16}{3}\cos^3\!\theta + C \;=\;-\frac{16}{3}\left(\frac{\sqrt{4-x^2}}{2}\right)^3 + C \;=\;-\frac{16}{3}\cdot\frac{\left(\sqrt{4-x^2}\right)^3}{8} + C$

. . . . . . . $= \;-\frac{2}{3}(4-x^2)^{\frac{3}{2}} + C$

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