For any integer n≥2, let ln=∫tannxdx. If ln=1atann-1x-bln-2for n≥2, then the ordered pair (a,b)=
n-1,(n-1)(n-2)
n-1,(n-2)(n-1)
(n,1)
(n-1,1)
Step1. The given integration :
ln=∫tannxdx
=∫tann-2xtan2xdx=∫tann-2x(sec2x-1)dx=∫(tann-2xsec2x-tann-2x)dx
Let tanx=t
⇒sec2xdx=dt
Now, ∫tann-2xsec2xdx=∫tn-2dt
=tn-2+1(n-2+1) ∵∫xndx=xn+1(n+1)
=tn-1(n-1)
Now, put the value of t then
=tann-1(n-1)
Step2. Find the ordered pairs (a,b):
And ln=∫tannxdx
So, ln=tann-1xn-1-ln-2
Comparing with ln=1atann-1x-bln-2then
a=n-1,b=1
Hence, the correct option is (D).