Incorrect integral

**thomas49th** · June 18th 2009, 06:45 AM

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

**malaygoel** · June 18th 2009, 06:54 AM

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

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

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

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

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

$=\frac{3a}{2}$

**malaygoel** · June 18th 2009, 06:57 AM

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

**thomas49th** · June 18th 2009, 06:58 AM

im an idiot...

how did they get there stupid $<br /> <br /> 3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0<br />$

thanks

**HallsofIvy** · June 18th 2009, 09:53 AM

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 $3a\left[\frac{1}{2}sin^2(x)\right]_0^{\pi/2}$ .

**thomas49th** · June 18th 2009, 10:26 AM

oh course the one of the products is the differential of the other. cheers

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