Question

f
dx.
sin² cos²x
equals
please tell i will mark as brainliest
​

Answer 1

Step-by-step explanation:

∫ sin²x cos²x dx =

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

I hope it helps...

Bye!