# [SOLVED] Quick easy integration question

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• February 2nd 2010, 07:29 AM
nunos
[SOLVED] Quick easy integration question
Is $\int \sin x \cos x = \frac{\sin^2 x}{2}$ ?

Thanks.
• February 2nd 2010, 07:30 AM
Henryt999
Integrate the product
Sure easy
$sinx*cosx = 0.5sin2x$
Can you take it from here?
• February 2nd 2010, 07:33 AM
nunos
Quote:

Originally Posted by Henryt999
Sure easy
$sinx*cosx = 0.5sin2x$

Perhaps you could you give a more detailed explanation... Thanks.
• February 2nd 2010, 07:36 AM
e^(i*pi)
Quote:

Originally Posted by nunos
Is $\int \sin x \cos x = \frac{\sin^2 x}{2}$ ?

Thanks.

Almost, you need a constant of integration though

$I = -\frac{1}{4}\cos (2x) + k = \frac{sin^2(x)}{2} + C$ where k is a constant and where $C = k - \frac{1}{4}$
• February 2nd 2010, 07:38 AM
Henryt999
Sure
$2sinx*cosx = sin2x$

$\frac{2*sinx*cosx}{2}$ = $0.5*sin(2x)$

And you know how to integrate that?
• February 2nd 2010, 07:46 AM
nunos
What I thought was using what I believe to be some sort of the "power rule".

When one does $\int x = \int x . x' = \frac{x}{2}$

so, one can apply that here:

$\int \sin x \cos x = \int \sin x (\sin x)' = \frac{\sin^2 x}{2}$

Is this thought process, correct, or have I made any mistake?

Thanks in advance.
• February 2nd 2010, 08:37 AM
HallsofIvy
Yes, that is correct. Let u= sin(x). Then du= cos(x) dx so your integral is
$\int sin(x)cos(x) dx= \int u du= \frac{1}{2}u^2+ C= \frac{sin^2(x)}{2}+ C$.

Note that you could also let u= cos(x). Then du= -sin(x)dx so your integral is
$\int sin(x)cos(x)dx= \int cos(x) (sin(x)dx)= -\int u du= -\frac{1}{2}u^2+ C'= -\frac{cos^2(x)}{2}+ C'$.

Since $cos^2(x)= 1- sin^2(x)$, $-\frac{cos^2(x)}{2}+ C'= -\frac{1}{2}+ \frac{sin^2(x)}{2}+ C'$. 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 $\frac{1}{2}\int sin(2x)dx$. Now let u= 2x so that du= 2 dx and dx= (1/2)du. Your integral is

$\frac{1}{2}\int sin(2x)dx= \frac{1}{2}\int u (\frac{1}{2}du)$ $= \frac{1}{4}\int sin(u)du= -\frac{1}{4}cos(u)+ C''= -\frac{1}{4}cos(2x)+ C''$.

But, now, $cos(2x)= cos^2(x)- sin^2(x)$ so that solution is the same as $\frac{1}{4}(cos^2(x)- sin^2(x))+ C''$. Replacing $sin^2(x)$ with $1- cos^2(x)$, that becomes $\frac{1}{4}(cos^2(x)- 1+ cos^(x))+ C''$ $= \frac{1}{4}(2 cos^2(x)- 1)+ C''= \frac{cos^2(x)}{2}+ C''- \frac{1}{4}$, the same as our second solution with $C'= C''- \frac{1}{4}$ and so the same as our first solution with $C= C''- \frac{3}{4}$.
• February 2nd 2010, 08:45 AM
nunos
Thanks HallsofIvy for your very clear explanation. I get it you know.