Is $\displaystyle \int \sin x \cos x = \frac{\sin^2 x}{2} $ ?
Thanks.
What I thought was using what I believe to be some sort of the "power rule".
When one does $\displaystyle \int x = \int x . x' = \frac{x}{2}$
so, one can apply that here:
$\displaystyle \int \sin x \cos x = \int \sin x (\sin x)' = \frac{\sin^2 x}{2}$
Is this thought process, correct, or have I made any mistake?
Thanks in advance.
Yes, that is correct. Let u= sin(x). Then du= cos(x) dx so your integral is
$\displaystyle \int sin(x)cos(x) dx= \int u du= \frac{1}{2}u^2+ C= \frac{sin^2(x)}{2}+ C$.
Note that you could also let u= cos(x). Then du= -sin(x)dx so your integral is
$\displaystyle \int sin(x)cos(x)dx= \int cos(x) (sin(x)dx)= -\int u du= -\frac{1}{2}u^2+ C'= -\frac{cos^2(x)}{2}+ C'$.
Since $\displaystyle cos^2(x)= 1- sin^2(x)$, $\displaystyle -\frac{cos^2(x)}{2}+ C'= -\frac{1}{2}+ \frac{sin^2(x)}{2}+ C'$. The two solutions are the same with C= C'- 1/2.
Henryt999 was pointing to a completely different but still valid solution. Since sin(2x)= 2sin(x)cos(x), sin(x)cos(x)= (1/2)sin(2x) and your integral becomes $\displaystyle \frac{1}{2}\int sin(2x)dx$. Now let u= 2x so that du= 2 dx and dx= (1/2)du. Your integral is
$\displaystyle \frac{1}{2}\int sin(2x)dx= \frac{1}{2}\int u (\frac{1}{2}du)$$\displaystyle = \frac{1}{4}\int sin(u)du= -\frac{1}{4}cos(u)+ C''= -\frac{1}{4}cos(2x)+ C''$.
But, now, $\displaystyle cos(2x)= cos^2(x)- sin^2(x)$ so that solution is the same as $\displaystyle \frac{1}{4}(cos^2(x)- sin^2(x))+ C''$. Replacing $\displaystyle sin^2(x)$ with $\displaystyle 1- cos^2(x)$, that becomes $\displaystyle \frac{1}{4}(cos^2(x)- 1+ cos^(x))+ C''$$\displaystyle = \frac{1}{4}(2 cos^2(x)- 1)+ C''= \frac{cos^2(x)}{2}+ C''- \frac{1}{4}$, the same as our second solution with $\displaystyle C'= C''- \frac{1}{4}$ and so the same as our first solution with $\displaystyle C= C''- \frac{3}{4}$.