 Quote by b0rsuk
1. The problem statement, all variables and given/known data
I'm unable to solve this integral. I get a result, but it doesn't match the solution.
\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}
[tex]\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}[/tex]
3. The attempt at a solution
(attachment?)
I'd be grateful for highlighting my errors.
|
After splitting this up into two integrals, the first one is fine.
You then made offsetting errors -- or simply had a typo.
[itex]\displaystyle -2\int\left(\sin\left(\frac{x}{2}\right)\cos\left( \frac{x}{2}\right)\right)dx=-4\int\left(\frac{1}{2}\sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\ right)\right)dx[/itex][itex]\displaystyle=-4\int\left(\sin(t)\cos(t)\right)dt[/itex]
After that you dropped a minus sign.
So, the final answer should be [itex]\displaystyle x-2\sin^2\left(\frac{x}{2}\right)+C[/itex]
I know that's
not the book's answer, but notice that
[itex]1-2\sin^2(\theta)=\cos(2\theta)[/itex]
So substituting x/2 for θ and doing a bit of algebra gives
[itex]\displaystyle -2\sin^2\left(\frac{x}{2}\right)=\cos(x)-1[/itex]
Therefore, [itex]\displaystyle x-2\sin^2\left(\frac{x}{2}\right)+C=x+\cos(x)-1+C[/itex]
-1+C is just a constant, one unit less than C.
So, your answer was correct except for a sign error.