Integral of Sqrt of (49-4x^2) and Differentiating Answer

1. Mar 17, 2009

ana111790

1. The problem statement, all variables and given/known data
$$\int\sqrt{49-4x^2}$$
I have done the integration to the best of my ability. My teacher requires that I differentiate the answer to get back to the original integrand. Can someone please check my work and help me along with the differentiation?

2. Relevant equations
U substitutions, and double angle formula ((sinx)^2=1/2 - (cosx)/2

3. The attempt at a solution

$$7\int\sqrt{1-(2x/7)^2}$$

cosu=2x/7
-sinu du=2/7 dx
u=arccos(2x/7)

Integral Rewritten as:

$$\int$$$$7\sqrt{1-cos^2}* -7(sinu)/2 du$$
$$7\sqrt{1-cos^2}= sinu$$ therefore:

=$$-49/2\int(sin^2u)du$$
=$$-49/2\int(1-cos^2)/2 du$$
=$$-49x/4 + 49(sinu)/2 + C$$
=$$-49x/4 + 49(sin(arccos(2x/7)))$$)/4 + C[/tex]

Differentiation:
$$-49/4 + [49(cos(arccos(2x/7))]/(4 \sqrt{1-(2x/7)^2}) * 4/49$$
=$$-49/4 + (2x)/(7\sqrt{49-4x^2})$$

Does this make sense at all?
How can I get to the original integrand given $$\int\sqrt{49-4x^2}$$

Last edited: Mar 17, 2009
2. Mar 17, 2009

gabbagabbahey

Careful, the square root sign always returns the positive root

$$\implies \sqrt{\sin^2(u)}=\left\{\begin{array}{lr}\sin(u), & \sin(u)\geq 0 \\ -\sin(u), & \sin(u)\leq 0\end{array}$$

For this reason, $\cos(u)=\frac{2x}{7}$ is a poor choice of substitution.

Here, there is another error; $$\int \cos^2(u) du \neq -2\sin(u)$$

3. Mar 17, 2009

ana111790

How do you suggest I solve it? I have literally spent days on this problem and I'm still lost :(

4. Mar 17, 2009

gabbagabbahey

Edit there is a little trick necesary, so show me what you get after applying integration by parts once, and I'll help you from there

Last edited: Mar 17, 2009
5. Mar 17, 2009

ana111790

I have tried Integration by Parts and it seems to never end.

I separated the integrand into 2 fractions, by multiplying top and bottom by sqrt(49-4x^2) to get:
49/ sqrt(49-4x^2) - 4x^2/(49-4x^2)
The 1st fraction I can integrate easily, the second I used Integration by Parts so that:

u=x^2
thus du=2x

dv=1/(1-(2x/7))
v=7/2 * arcsin(2x/7)

I kept doing IBP up to 5 times, and could not get an answer.
According to the book the answer should be xsqrt(49-4x^2) + 49/2 arcsin(2x/7) + C

6. Mar 17, 2009

gabbagabbahey

Hmmm... why not just use dv=dx and u=sqrt(49-4x^2)...what do you get after applying IBP once like this?

7. Mar 17, 2009

gabbagabbahey

Good .... now comes the tricky business; let's call your original integral $I$:

$$I\equiv \int \sqrt{49-4x^2)dx$$

What happens if you add and subtract 49 to the numerator of integral of [4x^2/(Sqrt(49-4x^2)]?

8. Mar 17, 2009

gabbagabbahey

Yes,

$$\int\frac{49}{\sqrt{49-4x^2}}dx=\int\frac{49}{7\sqrt{1-\left(\frac{2x}{7}\right)^2}}dx$$

You forgot to divide by that 7

9. Mar 17, 2009

gabbagabbahey

Welcome

10. Mar 17, 2009

xaos

hum... sqrt(sin^2(x))=abs(sin(x)). in trig substitutions, as far as i know, we almost always take the triangle to be in the first quadrant, so i dont think we should be so particular about this.

so sqrt(1-x^2)=cos(t) when x=sin(t) . here i prefer that x is vertical instead of horizontal.

cos^2(x)=(1+cos(2t))/2 and the result should follow. i think this was what the op was looking for.