# 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