Integrate:sinxcos(x-1/4.pi)dx
I guess maybe you mean
sin x cos(x-pi/4)dx
sin x cos y = .5( sin x +sin y) + .5 (sin x - sin y)
(remember "beat frequency" )
so
.5(sin x + sin(x-pi/4)dx
+
.5(sin x - sin(x-pi/4)dx
----------------------------
add (oh my)
sin x dx
-cos x + c
I guess maybe you mean
sin x cos(x-pi/4)dx
sin x cos y = .5 sin (x+y) + .5 sin (x-y)
(remember "beat frequency" )
so
.5sin ( 2x-pi/4)dx
+
.5sin (pi/4)dx
note sin pi/4 = sqrt2/2
-.25 cos (2x-pi/4) - .25 sqrt 2
note we no longer have sine curves with zero mean value but now have an average plus a cosine curve with twice the frequency.
-.25 cos (2x-pi/4) - .25 x sqrt 2 + c
To integrate the given expression ∫ sin(x)cos(x - 1/4π) dx, we can use the product-to-sum trigonometric identity:
sin(A)cos(B) = (1/2)[sin(A + B) + sin(A - B)]
Using this identity, we can rewrite the expression as:
∫ sin(x)cos(x - 1/4π) dx = ∫ (1/2)[sin(x + x - 1/4π) + sin(x - x + 1/4π)] dx
Simplifying further, we have:
∫ (1/2)[sin(2x - 1/4π) + sin(1/4π)] dx
Now, we can integrate each term separately:
∫ (1/2)sin(2x - 1/4π) dx + ∫(1/2)sin(1/4π) dx
The first integral can be solved by using a substitution. Let u = 2x - 1/4π, then du = 2 dx:
∫ (1/2)sin(2x - 1/4π) dx = (1/4) ∫ sin(u) du = -(1/4)cos(u) + C
Replacing u back with 2x - 1/4π, we have:
-(1/4)cos(2x - 1/4π) + C1
The second integral ∫(1/2)sin(1/4π) dx is a constant and can be pulled out of the integral:
(1/2)sin(1/4π) ∫ dx = (1/2)sin(1/4π)x + C2
Combining the two results, the final solution is:
-(1/4)cos(2x - 1/4π) + (1/2)sin(1/4π)x + C
Therefore, the integral of sin(x)cos(x - 1/4π) dx is -(1/4)cos(2x - 1/4π) + (1/2)sin(1/4π)x + C.