Register to reply

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

by b0rsuk
Tags: integral
Share this thread:
b0rsuk
#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
CodeCogsEqn.gif  
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Thaakisfox
#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
#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
#4
Dec26-11, 08:04 AM
HW Helper
Curious3141's Avatar
P: 2,952
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
#5
Dec26-11, 09:12 AM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Hello b0rsuk. Welcome to PF !

What is the correct answer?
b0rsuk
#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
#7
Dec26-11, 10:31 AM
HW Helper
Curious3141's Avatar
P: 2,952
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
#8
Dec26-11, 10:55 AM
Mentor
P: 15,170
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
#9
Dec26-11, 01:02 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
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.
Attached Thumbnails
CodeCogsEqn_gif_750x750_q85.jpg  


Register to reply

Related Discussions
Rewrite the integral as an equivalent iterated integral in the order Calculus & Beyond Homework 5
Using polar co-ord. to change double integral into single integral involving only r. Calculus & Beyond Homework 5
Is the ordinary integral a special case of the line integral? Calculus 3
Volume integral to spherical coords to contour integral Calculus & Beyond Homework 4
Is Cauchy's integral formula applicable to this type of integral? Calculus & Beyond Homework 4