Problem with Integration/Two Different Integrals

[yoshtov]

Homework Statement

int[sin(x)*cos(x)]dx

Homework Equations

U-substitution

The Attempt at a Solution

Alternative #1:

u=sin(x)
du=cos(x)dx

intdu = u^2/2 + K = sin^2(x)/2+K

Alternative #2

u=cos(x)
du=-sin(x)dx

-intdu = -u^2/2 + K = -cos^2(x)/2+K

From where I'm sitting, both approaches to using U-substitution are equally valid, but yield different solutions. As near as I can tell, they are not equal. Can you please point out where I made a logical misstep?

 

[Dick]

No mistake. They aren't equal. But they do differ by a constant. sin^2(x)=(-cos^2(x))+1. They both differentiate to sin(x)*cos(x).

 

[Char. Limit]

Let's rewrite your two Ks as K₁ and K₂

Then we can equate these two expressions to get:

[tex]\frac{sin^2(x)}{2} + K_1 = - \frac{cos^2(x)}{2} + K_2[/tex]

Then rewrite sin²(x) as 1-cos²(x) via the Pythagorean theorem, and we get...

[tex]\frac{1 - cos^2(x)}{2} + K_1 = - \frac{cos^2(x)}{2} + K_2[/tex]

Then, adding cos²(x)/2 to both sides, we get...

[tex]\frac{1}{2} + K_1 = K_2[/tex]

And since K₁ and K₂ are both arbitrary, there exists a K₁ and K₂ such that this is true. Therefore, the equations are really the same, up to a constant.

 

[yoshtov]

Remarkable! Thank you!