integral (sin x/2 - cos x/2)^2


by b0rsuk
Tags: integral
b0rsuk
b0rsuk is offline
#1
Dec26-11, 07:39 AM
P: 5
1. The problem statement, all variables and given/known data

I'm unable to solve this integral. I get a result, but it doesn't match the solution.

\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}
[tex]\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}[/tex]

3. The attempt at a solution

(attachment?)
I'd be grateful for highligting my errors.
Attached Thumbnails
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Thaakisfox
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#2
Dec26-11, 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]
Harrisonized
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#3
Dec26-11, 07:48 AM
P: 209
2sin(x)cos(x) = sin(2x)dx

∫ [2sin(x)] [cos(x)dx] = sin2(x)
∫ [-2cos(x)] [-sin(x)dx] = -cos2(x)
∫ sin(2x)dx = -(1/2)cos(2x)

Curious3141
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#4
Dec26-11, 08:04 AM
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integral (sin x/2 - cos x/2)^2


Quote Quote by b0rsuk View Post
1. The problem statement, all variables and given/known data

I'm unable to solve this integral. I get a result, but it doesn't match the solution.

\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}
[tex]\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}[/tex]

3. The attempt at a solution

(attachment?)
I'd be grateful for highligting my errors.
Hint: simplify with the double-angle formula first.
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.
SammyS
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#5
Dec26-11, 09:12 AM
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Hello b0rsuk. Welcome to PF !

What is the correct answer?
b0rsuk
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#6
Dec26-11, 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 arch-nemesis, math. This time, without time pressure. You rob me of any excuses I might have to procrastinate solving examples.
Curious3141
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#7
Dec26-11, 10:31 AM
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Quote Quote by b0rsuk View Post
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]\sin^2 t + \cos^2 t[/tex]
Does it equal 1 ?
--------------------------

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 arch-nemesis, math. This time, without time pressure. You rob me of any excuses I might have to procrastinate solving examples.
[itex]\sin^2x + \cos^2x = 1[/itex] is an identity. Meaning the value of x can be *anything*. Even if you replace the x with x/2, it still holds (just to be clear, that means [itex]\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} = 1[/itex]). Don't bother with substitution, etc. Just recognise this and reduce the integrand to 1 first.

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].
D H
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#8
Dec26-11, 10:55 AM
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P: 14,454
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 half-angle formulae [itex]2\sin^2(u/2) = 1-\cos u[/itex]. The half-angle 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]
SammyS
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#9
Dec26-11, 01:02 PM
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Quote Quote by b0rsuk View Post
1. The problem statement, all variables and given/known data

I'm unable to solve this integral. I get a result, but it doesn't match the solution.

\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}
[tex]\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}[/tex]

3. The attempt at a solution

(attachment?)
I'd be grateful for highlighting my errors.
After splitting this up into two integrals, the first one is fine.

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]
[itex]\displaystyle=-4\int\left(\sin(t)\cos(t)\right)dt[/itex]
After that you dropped a minus sign.

So, the final answer should be [itex]\displaystyle x-2\sin^2\left(\frac{x}{2}\right)+C[/itex]

I know that's not the book's answer, but notice that
[itex]1-2\sin^2(\theta)=\cos(2\theta)[/itex]

So substituting x/2 for θ and doing a bit of algebra gives

[itex]\displaystyle -2\sin^2\left(\frac{x}{2}\right)=\cos(x)-1[/itex]
Therefore, [itex]\displaystyle x-2\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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