Để tìm nguyên hàm của sin⁡(3x)cos⁡(5x)\sin(3x) \cos(5x)sin(3x)cos(5x) dx, ta sử dụng công thức sau:

sin⁡(A)cos⁡(B)=12[sin⁡(A+B)+sin⁡(A−B)]\sin(A) \cos(B) = \frac{1}{2}[\sin(A + B) + \sin(A - B)]sin(A)cos(B)=21​[sin(A+B)+sin(A−B)]

Trong trường hợp này, A = 3x và B = 5x. Vậy ta có:

sin⁡(3x)cos⁡(5x)=12[sin⁡(3x+5x)+sin⁡(3x−5x)]\sin(3x) \cos(5x) = \frac{1}{2}[\sin(3x + 5x) + \sin(3x - 5x)]sin(3x)cos(5x)=21​[sin(3x+5x)+sin(3x−5x)]

sin⁡(3x)cos⁡(5x)=12[sin⁡(8x)+sin⁡(−2x)]\sin(3x) \cos(5x) = \frac{1}{2}[\sin(8x) + \sin(-2x)]sin(3x)cos(5x)=21​[sin(8x)+sin(−2x)]

sin⁡(3x)cos⁡(5x)=12[sin⁡(8x)−sin⁡(2x)]\sin(3x) \cos(5x) = \frac{1}{2}[\sin(8x) - \sin(2x)]sin(3x)cos(5x)=21​[sin(8x)−sin(2x)]

Tiếp theo, ta tích phân.

∫sin⁡(3x)cos⁡(5x)dx=12∫[sin⁡(8x)−sin⁡(2x)]dx\int \sin(3x) \cos(5x) dx = \frac{1}{2}\int[\sin(8x) - \sin(2x)] dx∫sin(3x)cos(5x)dx=21​∫[sin(8x)−sin(2x)]dx

=12∫sin⁡(8x)dx−12∫sin⁡(2x)dx= \frac{1}{2}\int \sin(8x) dx - \frac{1}{2}\int \sin(2x) dx=21​∫sin(8x)dx−21​∫sin(2x)dx

=12⋅(−18cos⁡(8x))−12⋅(−12cos⁡(2x))+C= \frac{1}{2} \cdot (-\frac{1}{8} \cos(8x)) - \frac{1}{2} \cdot (-\frac{1}{2} \cos(2x)) + C=21​⋅(−81​cos(8x))−21​⋅(−21​cos(2x))+C

=−116cos⁡(8x)+14cos⁡(2x)+C= -\frac{1}{16} \cos(8x) + \frac{1}{4} \cos(2x) + C=−161​cos(8x)+41​cos(2x)+C

Vậy, nguyên hàm của sin⁡(3x)cos⁡(5x)\sin(3x) \cos(5x)sin(3x)cos(5x) là:

−116cos⁡(8x)+14cos⁡(2x)+C-\frac{1}{16} \cos(8x) + \frac{1}{4} \cos(2x) + C−161​cos(8x)+41​cos(2x)+C