please give the mistake in the following solution:
I=∫ (ax+b)sec x tan x dx4+tan2x dx (a,b>0) and lower limit 0 and upper limit π
splitting the numerator,
∫ax sec x tan x dx3+sec2x = ∫ a(π-x)sec x tan x3+sec2x dx = I1
so 2I1= aπ sec x tan x 3+sec2x
So I=(b+ (aπ/2)) dt3+t2 where t= sec x
integrating we have
I= (b+aÏ€/2)1√3tan-1 t√3
applying the limits from 1 to -1 as at x=0, t= 1 and at x=Ï€ ,t=-1
we have the answer as -Ï€(aÏ€+2b)6√3
but the answer given is Ï€(aÏ€+2b)3√3
actually in (0,Ï€), the function under the integral is always positive so how can the area under it's curve be negative ......so the negative sign is wrong............but i don't get the mistake in the above solution.......what went wrong ?
66The substitution t=sec x can not be applied because sec x is discontinuous at x=pi/2. Further the discontinuity is infinite in nature so this substitution won't help. you might consider breaking the limits in two parts: from 0 to pi/2 and then from pi/2 to pi.
