Question

∫[√(1 – sinx)/(1 + cosx)]e^(–x/2)dx = _________ + c 

(a) e^[(–x)/2]sec(x/2) 

(b) – e^[(–x)/2] sec(x/2) 

(c) – 2e^[(–x)/2] sec(x/2) 

(d) 2e^[(–x)/2] sec(x/4)

Answer 1

The correct option (b) – e^[(–x)/2] sec(x/2)

Explanation:

Let – (x/2) = t ➙        x = – 2t

∴ dx = – 2dt 

∴ I = ∫[√{1 – sin(– 2t)}/{1 + cos(– 2t)}] ∙ e^t(– 2dt) 

= ∫(– 2)e^tdt [√(1 + sin2t)/(1 + cos2t)] 

= ∫[(– 2)/(cos²t)] ∙ e^t √(cost + sint)dt 

= ∫[(– 1)/(cos²t)] ∙ e^t (cost + sint)dt 

= ∫ – e^t (+ sect + sect tant)dt 

= (– 1)∫e^t [f(t) + f'(t)]dt 

whose f(t) = sect and solution is given by e^tf(t) + c 

∴ I = e^t sect(– 1) + c 

= – e^tsect + c = – e^–(x/2)sec(x/2) + c