Question

integral sin²x.cos²x dx

Answer 1

Answer:

sin²x cos²x dx = (1/8)x - (1/8) sinx cos³x + (1/8) sin³x cosx + C

Step-by-step explanation:

Recall the double-angle identity sin(2x) = 2 sinx cosx;  

Hence:  

sinx cosx = (1/2) sin(2x) →  

sin²x cos²x = (sinx cosx)² = [(1/2) sin(2x)]² = (1/4)sin²(2x)  

Thus the given integral becomes:  

∫ sin²x cos²x dx = ∫ (1/4)sin²(2x) dx = (1/4) ∫ sin²(2x) dx  

Now you can reduce the order of the integrand using the half-angle identity:  

sin²x = (1/2) [1 - cos(2x)]  

and Therefore:  

sin²(2x) = (1/2) [1 - cos(4x)]  

Yielding:  

(1/4) ∫ (1/2) [1 - cos(4x)] dx =  

(1/4)(1/2) ∫ [1 - cos(4x)] dx =  

(1/8) ∫ [1 - cos(4x)] dx =  

Split it into:  

(1/8) ∫ dx - (1/8) ∫ cos(4x) dx =  

(1/8)x - (1/8) (1/4)sin(4x) + C =  

(1/8)x - (1/32)sin(4x) + C  

Then, taking it back in terms of sinx, cosx, you get (recall the double-angle identity cos(2x) = cos²x - sin²x, as well):  

∫ sin²x cos²x dx = (1/8)x - (1/32)sin(4x) + C →  

∫ sin²x cos²x dx = (1/8)x - (1/32) 2sin(2x) cos(2x) + C →  

∫ sin²x cos²x dx = (1/8)x - (1/16) 2sinx cosx (cos²x - sin²x) + C →  

∫ sin²x cos²x dx = (1/8)x - (1/8) (sinx cos³x - sin³x cosx) + C →  

Finally,  

∫ sin²x cos²x dx = (1/8)x - (1/8) sinx cos³x + (1/8) sin³x cosx + C