How can integrate "sin(p+q)x dx"?

Question

This a definite integral problem.How can integrate "sin(p+q)x dx"?
Give answers with explaination.

Answer 1

gen. rule:

integration of "sin(a)x dx" = -(cos(a)x )/a +c

So, ur answer is - (cos (p+q)x)/(p+q) +c

Answer 2

okay,
integrating sin(p+q)x dx = (- cos (p+q)x / p + q) + c
Breaking it down to simple terms,
if you were to integrate sin x, the result would be - cos x
if you were to integrate sin nx, then the expression of "x" is accompanied by an n and so the result would be - cos nx / n
In your question, n = p + q
So, you divide the whole expression by p + q
As there is no limit of your integration, don't forget to add the constant c.
If you still have not understood why this is so, then focus:
integration is the inverse of differenciation.
If you were to differenciate (- cos (p + q) / p + q) + c then the result would be:
d(- cos (p + q) x/ p + q) + c)/dx = sin (p + q) x * (p + q) / (p + q),
= sin (p + q) x
which is equal to the expression we have found earlier.
notice that we do not add 1 to the denominator as we would have if we were simply integrating (p + q)x, because it is a trigonometrical funtion, not a linear function.

Answer 3

This is in the form sin axcosbx+cosaxsinbx
Integrate both the termsby converting sin to cos or cos to sin
with the formula by parts
first function x integral of second function minus integral of differential of first function x integral of second function

Answer 4

since sin(p+q) is not affected by changes in x, we can consider sin(p+q) as a constant, a. so a = sin(p+q)
now our function is a*x dx
if we integrate this with resprect to x we get
a *x^2/2 + C
now substitute sin(p+q) back in the expression for a to get
sin(p+q)/2 * x^2

Answer 5

integral of sin(a)x dx is - cos(a)x/ a

similarly integral of sin(p+q) x dx= sin(px + qx) dx

integral of (sinpx cosqx + cos px sin qx) dx
integral 1/2(2 sin px cos qx ) dx
integral 1/2 ( sin(p+q)x + sin ( p-q)x ) dx
THEN U CAN APPLY SIMPLW INTEGRATION TO THIS I THINK

Answer 6

if (p+q)is a constant it is [cos(p+q)x/(p+q)] .now youapply the limits =>[1/(p+q)][cos(p+q)x1-cos(p+q)x2] where the limits are from 1 to 2