Integral from sin(x+y) from dx shows wierd result

I have to integrate the function f = sin (x+y) giveing that a=0 and b = pi /2 ( the heads of the integral ) ... the normal answer should had been sin(y) + cos(y) but my result is (2^(1/2)*sin(y + pi/4)) ... I think the problem are the heads becouse if I simply do int(f,x) I get the right answer , but I put the head int(f,x,a,b) it show that weird result

[Walter Roberson]

I get cos(y) + sin(y) when I try.

Can you post your code for testing ?

[Penescu Vlad]

yeah my bad is y not x *

Answer 1

[John D'Errico]

Computers don't really care that YOU think it looks wierd. The fact is, that form for the integral is correct, just as much so as the solution that you like. It may depend on your MATLAB release, of course, since with R2019b, I get:

syms x y
>> int(sin(x+y),x,0,pi/2)
ans =
cos(y) + sin(y)

However, if we recognize the trig identity:

sin(u + v) = sin(u)*cos(v) + cos(u)*sin(v)

Now consider what happens if u = y, and v = pi/4. Then

sin(y + pi/4) = sin(y)*cos(pi/4) + cos(y)*sin(pi/4)

But sin(pi/4) = cos(pi/4) = sqrt(2)/2. Therefore you should see that

cos(y) + sin(y) = 2/sqrt(2) * sin(y + pi/4) = sqrt(2)*sin(y + pi/4)

Same answer. You may think it is wierd. But at least your release of the symbolic toolbox decided it was a simple form to report the solution. In fact, it arguably is simpler, in the sense there is only one term in the solution. The solution you got was entirely valid, as was the alternative. Getting into the mind of a computer is sometimes difficult to know why it may choose any specific form for the solution.

[Penescu Vlad]

Ok I understand how it works now , but there is any way to show the exact answer sin(y) + cos(y) ? Becouse this is mainly for an exam I have and the other answer was not accepted ( even though was correct)... Can I make the computer think the result I want is the best one ?

[madhan ravi]

rewrite(answer, 'sincos')

[John D'Errico]

While rewrite is often a good idea, it does not seem to work in this case. I tried that myself before.

X = sqrt(sym(2))*sin(y + pi/4);
rewrite(X,'sincos')
ans =
2^(1/2)*sin(y + pi/4)

Sadly, rewrite seem to be often a bit arbitrary. On some problems it works wonders. On others it just laughs at me.

I suppose you could force things...

expand(subs(X,sin(y + pi/4),sin(y)*cos(pi/4) + cos(y)*sin(pi/4)))
ans =
cos(y) + sin(y)

It feels to be a bit of a kludge. I had to beat it upside the noggin to make it work there. And be careful, since MATLAB undoes all that work if you allow it to simplify:

simplify(cos(y) + sin(y))
ans =
2^(1/2)*sin(y + pi/4)

It really does think that to be a simpler form. Somedays we really want an unsimplify command. We have that, sort of, in a sense. You get the unsimplify from the 'all' property for simplify.

X
X =
2^(1/2)*sin(y + pi/4)

Xhat = simplify(X,5,'all',true)
Xhat =
 2^(1/2)*sin(y + pi/4)
       cos(y) + sin(y)
       
Xhat(2)
ans =
cos(y) + sin(y)

So simplify does see the one you like as ONE of the possible variations it could have chosen. Then we can extract the desired solution from the set of those equivalent forms it found. The list can get long if you tell it to look really hard.

simplify(X,50,'all',true)
ans =
                                                                                                                           2^(1/2)*sin(y + pi/4)
                                                                                                                           2^(1/2)*cos(y - pi/4)
                                                                                                                                 cos(y) + sin(y)
                                                                                                           2^(1/2) - 2*2^(1/2)*sin(y/2 - pi/8)^2
                                                                                                               2^(1/2)*(2*cos(y/2 - pi/8)^2 - 1)
                                                                                                              -2^(1/2)*(2*sin(y/2 - pi/8)^2 - 1)
                                                                                                     2^(1/2) + 2*2^(1/2)*(cos(y/2 - pi/8)^2 - 1)
                                                                                                                       sin(y) - 2*sin(y/2)^2 + 1
                                                                                                exp(-y*1i)*(1/2 + 1i/2) + exp(y*1i)*(1/2 - 1i/2)
 2^(1/2) - 2^(1/2)*cos(y/2)^2 - 2^(1/2)*sin(y/2)^2 + cos(y/2)^2 - sin(y/2)^2 + 2^(1/2)*cos(y/2)*sin(y/2)*(2 - 2^(1/2))^(1/2)*(2^(1/2) + 2)^(1/2)
                                                                                   2^(1/2)*(exp(- y*1i + (pi*1i)/4)/2 + exp(y*1i - (pi*1i)/4)/2)
                                                                         2^(1/2)*((exp(- y*1i - (pi*1i)/4)*1i)/2 - (exp(y*1i + (pi*1i)/4)*1i)/2)
                                                   2^(1/2) - 2*2^(1/2)*((exp(- (y*1i)/2 + (pi*1i)/8)*1i)/2 - (exp((y*1i)/2 - (pi*1i)/8)*1i)/2)^2
                                                      -2^(1/2)*(2*((exp(- (y*1i)/2 + (pi*1i)/8)*1i)/2 - (exp((y*1i)/2 - (pi*1i)/8)*1i)/2)^2 - 1)

Computers can be totally clueless some days. I do hope to get my copy of the mind-reading toolbox up and running. It is still in early beta test though, and the half the time it does work, then it tries to take over the world.

[madhan ravi]

+1.An impressive and detailed answer ;).

[Penescu Vlad]

I got another way to do it I think ... by using expand(int(f,x,a,b)) it shows the answer the way I wanted ... But thanks for the explacations, it helped a lot