Takehome Test Check

1. Sep 20, 2007

Lanza52

[SOLVED] Takehome Test Check

Have a take home test that is due tomorrow at noon. Got 3 of the 4 done and I thought that it would be a good idea to have better mathematicians then myself check it for me. So, if anybody feels like doing the work;

1. Integration by Parts

$$\int\sin(log_{2}x)dx$$

$$u = \sin(log_{2}x)$$

$$du = \frac{\cos(log_{2}x)}{xlog2}dx$$

dv = dx, v = x

$$x\sin(log_{2}x)-\int\frac{x\cos(log_{2}x)}{xlog2}$$

$$x\sin(log_{2}x)-\frac{1}{log2}\int\cos(log_{2}x)dx$$

Then do it again.

$$u = \cos(log_{2}x)$$

$$du = \frac{-\sin(log_{2}x)}{xlog2}dx$$

dv = dx, v = x

$$x\cos(log_{2}x)+\int\frac{x\sin(log_{2}x)}{xlog2}$$

$$x\cos(log_{2}x)+\frac{1}{log2}\int\sin(log_{2}x)dx$$

Now put it all together.

$$x\sin(log_{2}x)-\frac{1}{log2}(x\cos(log_{2}x)+\frac{1}{log2}\int\sin(log_{2}x)dx)=\int\sin(log_{2}x)dx$$

$$x\sin(log_{2}x)-\frac{1}{log2}x\cos(log_{2}x)=\frac{log^{2}2+1}{log^{2}2}\int\sin(log_{2}x)dx$$

$$\frac{log^{2}2(\sin(log_{2}x)-\frac{1}{log2}x\cos(log_{2}x))}{log^{2}2+1}=\int\sin(log_{2}x)dx$$

Last edited: Sep 20, 2007
2. Sep 20, 2007

Dick

You were doing fine until you tried to "put it all together". Then there's some algebraic slip ups. Try it again. E.g. I get (log2)^2+1 in the denominator.

3. Sep 20, 2007

Lanza52

Ah, yea, slight of...latex? Just got impatient while painstakingly going through the code. 1/Log2 * 1/log2 = 1/(log2)^2......(log2)^2/(log2)^2 - 1/(log2)^2 = (log2)^2-1 / (log2)^2 and from there multiply both sides by denom then devide both sides by num.

Correct now?

4. Sep 20, 2007

Dick

That looks familiar - but don't you get ((log2)^2+1)/(log2)^2??? In 'putting it all together' you have a sign error on integral(sin) on the LHS.

5. Sep 20, 2007

Lanza52

Edited the original post. How's that?

6. Sep 20, 2007

Dick

I don't see any change. Did the TeX not regenerate?

7. Sep 20, 2007

Lanza52

Got to push refresh button.

8. Sep 20, 2007

Dick

Ok, got it. You left out an 'x' in the sin term, but that's just a typo.

9. Sep 20, 2007

Lanza52

Thanks again =P