Question

int dx/sin²x . cos²x
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Answer 1

Answer:

Now, sin²x cos²x

= 1/4 (4 sin²x cos²x)

= 1/4 (2 sinx cosx)²

= 1/4 (sin2x)²

= 1/4 sin²2x

Now, ∫ dx/(sin²x cos²x)

= 4 ∫ dx/sin²2x

= 4 ∫ cosec²2x dx

= - 4 ∫ (- cosec²2x) dx

= -4/2 ∫ d (cot2x)

= - 2 cot2x + c, where c is integral constant

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Answer 2

Answer: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

Step-by-step explanation: