Question

∫[√(tanx) + √(cotx)]dx = √2 sin^–1(______) + c 

(a) sinx – cosx 

(b) cosx – sinx 

(c) sin(x/2) – cos(x/2) 

(d) cos(x/2) – sin(x/2)

Answer 1

The correct option (a) sinx – cosx  

Explanation:

given: √(tanx) + √(cotx) 

= [(sinx + cosx)/√(sinx ∙ cosx)] 

= [{√2(sinx + cosx)}/√{1 – (1 – 2sinxcosx)}] 

= [{√2(sinx + cosx)}/√{1 – (sinx – cosx)²}] 

Let sinx – cosx = t 

∴ (cosx + sinx)dx = dt 

∴ I = ∫√(tanx) + √(cotx)dx 

= ∫[{√(2)dt}/√(1 – t²)] 

= √2∫[dt/√(1 – t²)] 

= √(2) sin^–1t + c 

= √(2) sin^–1(sinx – cosx) + c