∫(cos 4x-1)/(cot x-tan x) dx is equal to

Option 1 - -1/2 cos 4x + C

Option 2 - -1/4 cos 4x + C

Option 3 - -1/2 sin 2x + C

Option 4 - 1/4 ln |cos 2x| - (cos² 2x)/4 + C

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Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 4

Detailed Solution:

I = ∫ (cos 4x-1)/ (cot x-tan x) dx = ∫ (2 cos² 2x-1-1)/ (cos²x-sin²x)/ (sin x cos x) dx
= 2 ∫ (cos² 2x-1) sin x cos x) / (cos 2x) dx = ∫ (cos² 2x-1) sin 2x) / (cos 2x) dx
Put cos 2x = t
-2 sin 2x dx = dt
I = -1/2 ∫ (t²-1)/t dt = 1/2 ∫ (1/t - t) dt
= 1/2 (ln|t| - t²/2) + c
= 1/2 ln|cos 2x| - 1/4 cos² (2x)

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