# Incorrect integral

• Jun 18th 2009, 07:45 AM
thomas49th
Incorrect integral
Hi, I was doing a question with this simple integral

$3a\int^{\frac{\pi}{2}}_0cosxsinxdx$

so rather than do it by parts (which probably would turn out to be double back on itself so you get the integral again), I opted for using sin2x = 2sinxcosx

so that means the integral simplified to

$3a\int^{\frac{\pi}{2}}_0\frac{1}{2}sin2x$

which integrates to

$3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0$

so wack in the limits and I get 3a[1-1] = 0

but that isn't right is it?

The answer scheme suggest $3a\int^{\frac{\pi}{2}}_0cosxsinxdx$ = $3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0$

but i) How come my method doesn't work
ii) How can the mark scheme do that. You can't just do that!?

Thanks :)
• Jun 18th 2009, 07:54 AM
malaygoel
Quote:

Originally Posted by thomas49th

$3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0$

so wack in the limits and I get 3a[1-1] = 0

$3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0$

$=-\frac{3a}{4}[cos2x]^{\frac{\pi}{2}}_0$

$=-\frac{3a}{4}[cos{\pi}-cos(0)]$

$=-\frac{3a}{4}[-1-1]$

$=\frac{3a}{2}$
• Jun 18th 2009, 07:57 AM
malaygoel
Quote:

Originally Posted by thomas49th

The answer scheme suggest $3a\int^{\frac{\pi}{2}}_0cosxsinxdx$ = $3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0$

ii) How can the mark scheme do that. You can't just do that!?

Thanks :)

It is incorrect

it should be
$3a[\frac{1}{2}sin^{2}x]^{\frac{\pi}{2}}_0$
• Jun 18th 2009, 07:58 AM
thomas49th
im an idiot...

how did they get there stupid $

3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0
$

thanks :)
• Jun 18th 2009, 10:53 AM
HallsofIvy
The simplest way to do $\int sin(x)cos(x) dx$ is neither "by parts" or using a trig identity like that. Let u= sin(x) so that du= cos(x) dx and this becomes the simple integral $\int u du$.

An integral of that is $\frac{1}{2} u^2$. That's where they get $3a\left[\frac{1}{2}sin^2(x)\right]_0^{\pi/2}$.
• Jun 18th 2009, 11:26 AM
thomas49th
oh course the one of the products is the differential of the other. cheers :)