What method do I use to calculate the integral of sin (x) * sin (nx) dx?

Question

I want to find the integral of [sin (x) * sin (nx)] dx. If I use substitution for nx, how would I account for the 'x' in the first sine term? Would I use integration by parts?

Answer 1

Start by letting I = integral of [sin(x)*sin(nx)]
Use integration by parts twice.
Let u = sin(x) and dv = sin(nx)
After using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.
(Your instructor should have shown you this trick)

Answer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]
I did this quickly, so you should check my answer by working it out.

Answer 2

You can also use the identity
2sin(x)sin(y)=
cos(x-y)-cos(x+y),
so sin(x)sin(nx)=
(1/2)[cos((n-1)x)-cos((n+1)x)]
The integral is then easy by two substitutions.

Answer 3

You don't use chain rule to integrate. Chain rule is used for derivation. Integration by parts sounds like a logical choice.
Here is what I came up with:
int [sin(x)*sin(nx)],
let u = sin(x)
du = con(x)dx
dv = sin(nx) dx
v = -1/n cos(nx)
using integration by parts formula,
-1/n cos(nx)*sin(x)+int [1/n cos(nx)* cos(x) dx ]
let's integrate second part of above,
let u = cos(x)
du = -sin(x)dx
dv = cos(nx) dx
v = 1/n sin(nx)
using integration by parts formula again for the second part,
-1/n cos(nx)*sin(x)+1/n^2[sin (nx)*cos(x)- int sin(nx)*sin(x) dx]
1/n^2 int [sin(nx)*sin(x)dx] = -1/n cos(nx)*sin(x)+1/n^2[sin (nx)cos(x)]
int [sin(nx)*sin(x)dx] = -n cos(nx)*sin(x)+[sin (nx)*cos(x)]

Please double check my answer.

Answer 4

I think integration by parts is your best bet or theirs always chain rule.

Answer 5

i did no longer verify your answer, yet greater beneficial attitude could be as decrease than. cos(2x) sin(x) = (a million/2) * 2cos(2x)sin(x) = (a million/2) [sin(2x + x)/2 - sin(2x - x)/2] = (a million/2) [sin(3x)/2 - sin(x/2)] => imperative = (a million/2) [ - (2/3) cos(3x/2) + 2 cos(x/2)] + c = - (a million/3) cos(3x/2) + cos(x/2) + c.