# Math Help - triggy substitution

1. ## triggy substitution

find the following integral:

$\int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx$

what i was thinking was:
$x = \sqrt{3}tan u$
$dx = \sqrt{3}sec²u du$

then i get:

$\int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra c{5}{2}}} \cdot \sqrt{3} sec^2u du$

is this right?

2. Originally Posted by qzno
find the following integral:

$\int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx$

what i was thinking was:
$x = \sqrt{3}tan u$
$dx = \sqrt{3}sec²u du$

then i get:

$\int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra c{5}{2}}} \cdot \sqrt{3} sec^2u du$

is this right?
Yep - keep going!

3. does that equal:

$\int^3_0 \frac{tan^3u}{\sqrt{3} sec^{\frac{5}{2}}u} du$

????

4. Originally Posted by qzno
does that equal:

$\int^3_0 \frac{tan^3u}{\sqrt{3} sec^{\frac{5}{2}}u} du$

????
No! $\int^{\pi/3}_0 \frac{\tan^3u \sec^2 u}{\sqrt{3} \sec^{5}u} du = \frac{1}{\sqrt{3}} \int^{\pi/3}_0 \frac{\tan^3u }{ \sec^{3}u} du$ Note - the limits change.

5. why is it from 0 to pi/3 now?

6. Originally Posted by qzno
why is it from 0 to pi/3 now?
You original limits were where x was equal to 3 and 0, you are now however working with u.

Using your subsitution $x = \sqrt{3}tan u$, rearrange to get:

$u= arctan \frac{x}{\sqrt{3}}$

From this, sub in your two values of x, you then get the new values that danny mentioned.

7. Originally Posted by qzno
why is it from 0 to pi/3 now?
The original problem was $x = 0\; \text{to}\, x = 3$. Introducing the substitution $x = \sqrt{3} \tan u$ meanins that when you switch the problem in terms of u, the limits of integration will change. So, how do you find new limits - use the substitution. In your case

$x = 0\;\;\; \Rightarrow \;\;\; \sqrt{3} \tan u = 0\;\;\; \Rightarrow\; u = 0$
$x = 3\;\;\; \Rightarrow \;\;\; \sqrt{3} \tan u = 3\;\;\; \Rightarrow\; u = \pi/3.$

8. ah i understand! : ) thanks!

ill be posting my answer shortly hopefully

9. i got my final answer to be:

$\frac{5}{24\sqrt{3}}$

10. Originally Posted by qzno
i got my final answer to be:

$\frac{5}{24\sqrt{3}}$

11. Originally Posted by qzno
find the following integral:

$\int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx$

what i was thinking was:
$x = \sqrt{3}tan u$
$dx = \sqrt{3}sec²u du$

then i get:

$\int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra c{5}{2}}} \cdot \sqrt{3} sec^2u du$

is this right?
alternatively you can put $3+x^2=u^2.$ then $xdx=udu$ and so your integral becomes: $\int_{\sqrt{3}}^{2 \sqrt{3}} \frac{u^2 - 3}{u^4} \ du$ and you know how to finish this!

12. Hello,

Another method, not necessarily more beautiful...
$\int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx$

Let $t=\frac{1}{(3+x^2)^{3/2}}$
$dt=\frac{dx \cdot 2x \cdot (-3/2)}{(3+x^2)^{5/2}}=\frac{-3x ~dx}{(3+x^2)^{5/2}}$

$x=0 \Rightarrow t=3^{-3/2}=a$
$x=3 \Rightarrow t=12^{-3/2}=b$

And it follows that $3+x^2=t^{-2/3} \Rightarrow x^2=t^{-2/3}-3$

Hence the integral is now :
$\int_a^b \frac{x \cdot x^2}{(3+x^2)^{5/2}} \cdot \frac{(3+x^2)^{5/2}}{-3x} ~dt=-\frac 13 \int_a^b x^2 ~dt=-\frac 13 \int_a^b t^{-2/3}-3 ~dt$

Okay...