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Integral (sin x/2  cos x/2)^2by b0rsuk
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#1
Dec2611, 07:39 AM

#2
Dec2611, 07:45 AM

P: 263

You didnt do the substitution properly.
And there is no need for substitution. Note the identity: [itex]\sin(2x)=2\sin x \cos x[/itex] 


#3
Dec2611, 07:48 AM

P: 209

2sin(x)cos(x) = sin(2x)dx
∫ [2sin(x)] [cos(x)dx] = sin^{2}(x) ∫ [2cos(x)] [sin(x)dx] = cos^{2}(x) ∫ sin(2x)dx = (1/2)cos(2x) 


#4
Dec2611, 08:04 AM

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Integral (sin x/2  cos x/2)^2
sin(2z) = 2(sinz)(cosz) Now put z = (1/2)x and see what you get on the RHS. What do you get on the LHS? Integrate that. 


#5
Dec2611, 09:12 AM

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Hello b0rsuk. Welcome to PF !
What is the correct answer? 


#6
Dec2611, 10:26 AM

P: 5

Hmm. In such case,
[tex]2 \int \sin \frac{x}{2} \cos \frac{x}{2} dx = \int 2 \sin \frac{x}{2} \cos \frac{x}{2} dx = \int \sin x dx = \cos x + C_1[/tex] But what about the first integral ? I know: [tex]\sin^2 x + \cos^2 x = 1[/tex] But I have: [tex]\int\sin^2 \frac{x}{2} + \cos^2\frac{x}{2} dx[/tex] Can I simply get around that with substitution ? Say, [tex]\frac{x}{2} = t, \frac{1}{2} = dt[/tex] Then I get [tex]\frac{1}{2}(\sin^2 t + \cos^2 t)[/tex] Does it equal 1 (or, actually, 1/2) ?  The solution, according to the book: [tex]x + \cos x + C[/tex] Yes, that's a plus.  You people are scary. I'm a CS graduate and I came here to defeat my archnemesis, math. This time, without time pressure. You rob me of any excuses I might have to procrastinate solving examples. 


#7
Dec2611, 10:31 AM

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So the first integrand is 1, and the integral is x. The second integrand is [itex]\sin x[/itex] (you correctly reflected the minus sign in your first post, but then you must have forgotten it in your most recent one). The integral is [itex]\cos x[/itex]. So the final answer is [itex]x + \cos x + C[/itex]. 


#8
Dec2611, 10:55 AM

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P: 15,155

In this case, simple trig substitutions are all you need.
[tex]\left(\sin\frac x 2  \cos \frac x 2\right)^2 = 1  \sin x[/tex] This is an identity, so [itex]\int (\sin x/2  \cos x/2)^2 dx = \int (1\sin x)\,dx[/itex]. The right hand side obviously integrates to [itex]x+\cos x + c[/itex]. So how to arrive at that identity? One way is to expand the square, [tex]\left(\sin\frac x 2  \cos \frac x 2\right)^2 = \sin^2 \frac x 2  2 \sin \frac x 2 \cos \frac x 2 + \cos^2 \frac x 2[/tex] Now use the identities [itex]\sin^2 \theta + \cos^2\theta = 1[/itex] and [itex]\sin(2\theta) = 2 \sin\theta\cos\theta[/itex] to arrive at the desired identity [tex]\left(\sin\frac x 2  \cos \frac x 2\right)^2 = 1  \sin x[/tex] Another way is to use the identity [itex]\sin\theta \cos \theta = \sqrt 2 \sin(\theta\pi/4)[/itex]. This comes in handy at times, but doesn't quite fall into the camp of identities you should just "know". With this, [tex]\left(\sin\frac x 2  \cos \frac x 2\right)^2 = 2\sin^2\left(\frac x 2  \frac \pi 4\right) = 2\sin^2\left(\frac 1 2(x\pi/2)\right)[/tex] Now use the halfangle formulae [itex]2\sin^2(u/2) = 1\cos u[/itex]. The halfangle formulae come into play quite often. With this, [tex]\left(\sin\frac x 2  \cos \frac x 2\right)^2 = 1\cos\left(x\frac{\pi} 2\right) = 1  \sin x[/tex] 


#9
Dec2611, 01:02 PM

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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]After that you dropped a minus sign.[itex]\displaystyle=4\int\left(\sin(t)\cos(t)\right)dt[/itex] So, the final answer should be [itex]\displaystyle x2\sin^2\left(\frac{x}{2}\right)+C[/itex] I know that's not the book's answer, but notice that [itex]12\sin^2(\theta)=\cos(2\theta)[/itex]Therefore, [itex]\displaystyle x2\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. 


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