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

[b0rsuk]

Homework Statement

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}
\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}

The Attempt at a Solution

(attachment?)
I'd be grateful for highligting my errors.

 

[Thaakisfox]

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

 

[Harrisonized]

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]

Homework Statement

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}
\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}

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]

Hello b0rsuk. Welcome to PF !

What is the correct answer?

 

[b0rsuk]

Hmm. In such case,
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

But what about the first integral ? I know:
\sin^2 x + \cos^2 x = 1
But I have:
\int\sin^2 \frac{x}{2} + \cos^2\frac{x}{2} dx
Can I simply get around that with substitution ? Say,
\frac{x}{2} = t, \frac{1}{2} = dt
Then I get
\frac{1}{2}(\sin^2 t + \cos^2 t)
Does it equal 1 (or, actually, 1/2) ?
--------------------------

The solution, according to the book:
x + \cos x + C
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]

Hmm. In such case,
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

But what about the first integral ? I know:
\sin^2 x + \cos^2 x = 1
But I have:
\int\sin^2 \frac{x}{2} + \cos^2\frac{x}{2} dx
Can I simply get around that with substitution ? Say,
\frac{x}{2} = t, \frac{1}{2} = dt
Then I get
\sin^2 t + \cos^2 t
Does it equal 1 ?
--------------------------

The solution, according to the book:
x + \cos x + C
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.

\sin^2x + \cos^2x = 1 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 \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} = 1). 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 -\sin x (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 \cos x.

So the final answer is x + \cos x + C.

 

[D H]

In this case, simple trig substitutions are all you need.
\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 1 - \sin x
This is an identity, so \int (\sin x/2 - \cos x/2)^2 dx = \int (1-\sin x)\,dx. The right hand side obviously integrates to x+\cos x + c.

So how to arrive at that identity?One way is to expand the square,
\left(\sin\frac x 2 - \cos \frac x 2\right)^2 =<br /> \sin^2 \frac x 2 - 2 \sin \frac x 2 \cos \frac x 2 + \cos^2 \frac x 2
Now use the identities \sin^2 \theta + \cos^2\theta = 1 and \sin(2\theta) = 2 \sin\theta\cos\theta to arrive at the desired identity
\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 1 - \sin x

Another way is to use the identity \sin\theta- \cos \theta = \sqrt 2 \sin(\theta-\pi/4). This comes in handy at times, but doesn't quite fall into the camp of identities you should just "know". With this,
\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)
Now use the half-angle formulae 2\sin^2(u/2) = 1-\cos u. The half-angle formulae come into play quite often. With this,
\left(\sin\frac x 2 - \cos \frac x 2\right)^2 = 1-\cos\left(x-\frac{\pi} 2\right) = 1 - \sin x

 

[SammyS]

Homework Statement

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}
\int(sin \frac{x}{2} - cos \frac{x}{2})^2 \mathrm{dx}

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.

\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

\displaystyle=-4\int\left(\sin(t)\cos(t)\right)dt​

After that you dropped a minus sign.

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

I know that's not the book's answer, but notice that

1-2\sin^2(\theta)=\cos(2\theta)

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

\displaystyle -2\sin^2\left(\frac{x}{2}\right)=\cos(x)-1​

Therefore, \displaystyle x-2\sin^2\left(\frac{x}{2}\right)+C=x+\cos(x)-1+C

-1+C is just a constant, one unit less than C.

So, your answer was correct except for a sign error.