Question

∫[(cotxdx)/{√(cos⁴x + sin⁴x)}] = _________ + c 

(a) (1/2)log|cot²x + √(cot⁴ + 1)| 

(b) – (1/2)log|cot²x + √(cot⁴x + 1)|

(c) (1/2)log|tan²x + √(tan⁴ + 1)|

(d) – (1/2)log|cotx + √(cot⁴ + 1)|

Answer 1

The correct option (b) – (1/2)log|cot²x + √(cot⁴x + 1)|  

Explanation:

[(cotx)/√(cos⁴x + sin⁴x)] = [(cotx ∙ cosec²x)/√(1 + cot⁴x)] 

Let cot²x = t 

∴ 2cotx(– cosec²x)dx = dt 

∴ I = ∫[{–(dt)}/√(1 + t²)] 

= – (1/2)∫[dt/{√(1 + t²)}] 

= – (1/2)log|t + √(1 + t2)| + c 

= – (1/2)log|cot²x + √(1 + cot⁴x)| + c