Have you seen this proof before?

Question

I need help to prove this please.
Integrate(from pi to pi) of sin(mx) sin(nx) dx is pi if n=m and the integration equals zero is n not equals to m.
Yes, it is integrating from +pi to +pi.
Please show me the proves. You can show either one. Thank you!

Answer 1

If you don't mean the integration from -pi to pi then the proof is trivial, otherwise:

int sin(mx)sin(nx) dx , m does not equal n
= [-cos(mx)sin(nx)/m] + (n/m)* int cos(mx)*cos(nx)
= (n/m)*( [sin(mx)cos(nx)/n] + (n/m)*int sin(nx)sin(mx) )
(1 - (n/m)^2)*int sin(mx)sin(nx) dx = 0
And just divide through to complete the proof. From the second line we get the same result for cos(nx)cos(mx).
If m=n then
int sin^2(nx) dx = int (1 - cos(2nx))/2 dx
= [x/2 - sin(2nx)/4n ] = pi

Answer 2

Integrating from A to B gives you an answer unless A is the same value as B then the answer will be zero.

Integrating, your taking the area under the curve and if you are not taking any area then of course you get a zero.

Edit:

Its like doing this..... plug y = x on your calculator.

When you take the integral of y = x from say.... 0 to 2 you get 2 as your answer.

If you look at the graph, its a triangle where the equation of its area is (1/2)(base)(height)

The base in this case is 2, because we went from 0 to 2. the height is 2 because when you have y = x and plug in 2 for x, you also get 2 for y.

Now you know that the base and height are both equal to 2, you can find the are which is (1/2)(2)(2) and that equals to 2.

Now if you integrated from 2 to 2, then your base is equal to zero. So.... (1/2)(0)(2) and your answer is 0.

Answer 3

There seems to be a misprint in your problem.
The integral from pi to pi of anything is 0.
Maybe you mean -pi to pi.
Please clarify!