Question

If ∫[dx/(sin6x + cos6x)] = K tan^–1[(tan2x)/2] + c then K = _______

(a) (1/2) 

(b) – 1 

(c) 1 

(d) – (1/2)

Answer 1

The correct option (c) 1

Explanation:

I = ∫[dx/(sin⁶x + cos⁶x)] 

= ∫[1/(1 – 3sin²xcos²x)]dx 

= ∫[1/{1 – (3/4)(4sin²xcos²x)}]dx 

= ∫[4/(4 – 3sin²2x)]dx 

= ∫[(4sec²2x)/(4sec²2x – 3tan²2x)]dx 

= ∫[(4sec²2x dx)/{4(1 + tan²2x) – 3tan²2x}]dx 

= ∫[(4sec²2x dx)/(4 + tan²2x)] 

take tan2x = t 

∴ sec²2x dx(2) = dt 

∴ I = ∫[(2 dt)/(4 + t²)] 

= 2∫[dt/(t² + 2²)] 

= 2 × (1/2) tan^–1(t/2) + c 

I = tan–1[(tan2x)/2] + c 

Comparing with 

I = K tan^–1[(tan2x)/2] + c 

∴ K = 1