prove the following integrals:
a)sin3xcos7xdx = -1/20cos(10x)+1/8cos(4x)
i know i have asked this question before but i am very confused bobpursley, can you please show me the steps of proves.
thanks in advanced
This is all trig.
Sum formulas for sine and cosine
sin (s + t) = sin s cos t + cos s sin t
cos (s + t) = cos s cos t – sin s sin t
Use those formulas on the left, simplify, then do the integral. Is is mostly messy algebra, and a minimal amount of calculus.
To prove the integral ∫ sin(3x)cos(7x) dx = -1/20cos(10x) + 1/8cos(4x), we can use the trigonometric identity:
sin(A)cos(B) = 1/2[ sin(A - B) + sin(A + B) ]
Let's rewrite the original integral using this identity:
∫ sin(3x)cos(7x) dx = 1/2 ∫ [ sin(3x - 7x) + sin(3x + 7x) ] dx
Simplifying further:
∫ sin(3x)cos(7x) dx = 1/2 ∫ [ sin(-4x) + sin(10x) ] dx
Now, let's solve each integral separately:
∫ sin(-4x) dx = -1/4 cos(-4x) + C = -1/4 cos(4x) + C
∫ sin(10x) dx = -1/10 cos(10x) + C
Substituting the values back into the original equation:
∫ sin(3x)cos(7x) dx = 1/2 [ -1/4 cos(4x) + -1/10 cos(10x) ] + C
Simplifying further:
∫ sin(3x)cos(7x) dx = -1/8 cos(4x) - 1/20 cos(10x) + C
We can confirm that this is the correct integral solution by deriving it:
d/dx [ -1/8 cos(4x) - 1/20 cos(10x) + C ]
= 1/2 sin(4x) + 1/2 sin(10x)
Using the double angle formulas:
= 1/2 [2 sin(2x)cos(2x)]/2 + 1/2 [2 sin(5x)cos(5x)]/2
= sin(2x)cos(2x) + sin(5x)cos(5x)
Using the identity sin(A)cos(B) = 1/2[ sin(A - B) + sin(A + B) ]:
= 1/2 [sin(4x) + sin(0)] + 1/2 [sin(10x) + sin(0)]
= 1/2 sin(4x) + 1/2 sin(10x)
This confirms that the integral we found is correct:
∫ sin(3x)cos(7x) dx = -1/20cos(10x) + 1/8cos(4x) + C