Question

The value of integral ∫[2/(sec x + cosecx + tanx + cot x)]dx for x ∈ [0, π/4] is........ 

(a) 0 

(b) 1 – (π/4) 

(c) (π/4) + 1 

(d) (π/2) + 1

Answer 1

The correct option (b) 1 – (π/4)   

Explanation:

[2/(secx + cosecx + tanx + cotx)] 

= [2/{[1/(cos x)] + [1/(sin x)] + [(sin x)/(cos x)] + [(cos x)/(sin x)]}] 

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

= [(2 sinx ∙ cosx)/(sinx + cosx + 1)] 

= [{2sinx ∙ cosx × [1/(cos x)]}/(tanx + 1 + secx)] 

= [(2 sinx)/(tanx + 1 + secx)] 

= [{2sinx(tanx + 1 – secx)}/{(tanx + 1 + secx)(tanx + 1 – secx)}] 

= [{2sin x(tanx + 1 – secx)}/{(tan x + 1)² – sec²x}] 

= [{2sinx(tanx + 1 – secx)}/(tan²x + 2tanx + 1– sec²x)] 

= [{2sinx(tanx + 1 – secx)}/(2tanx)] 

= sinx + cosx – 1 

∴ I = ^(π/4)∫₀ (sinx + cosx – 1)dx 

= [– cosx + sinx – x]₀^(π/4) 

= – [cos(π/4) – cos0] + [sin(π/4) – sin0] – [(π/4) – 0] 

= – (1/√2) + 1 + (1/√2) – (π/4) 

= 1 – (π/4).