Answer: Let I n =∫ (secx+tanx) n sec 2 x dx (for n>1) ∴I n =∫ ( 1+tan 2 x +tanx) n sec 2 x dx (as sec 2 θ=1+tan 2 θ) Now, let t=tanx⇒dt=sec 2 xdx, we then have I n =∫ ( 1+t 2 +t) n 1 dt Making a hyperbolic substitution t=sinhy⟹dt=coshydy , we have I n =∫ ( 1+sinh 2 y +sinhy) n coshy dy Now, as sinhy= 2 e y −e −y and 1+sinh 2 y=cosh 2 y, we get I n = 2 1 ∫ e ny e y +e −y dy ∴I n = 2 1 ∫[e −(n−1)y +e −(n+1)y ]dy ∴I n =− 2(n−1) e −(n−1)y − 2(n+1) e −(n+1)y +C Now, re-substitute coshy−sinhy=e −y and sinhy=tanx, coshy=secx, we have I n =− 2(n−1) (secx−tanx) (n−1) − 2(n+1) (secx−tanx) (n+1) +C Applying the limits, we get I n =[− 2(n−1) (secx−tanx) (n−1) − 2(n+1) (secx−tanx) (n+1) ] 0 π/2 ∴I n = n 2 −1 n (Note that lim x→∞ secx−tanx=0) Hence, proved Reply
Answer:
Let I
n
=∫
(secx+tanx)
n
sec
2
x
dx (for n>1)
∴I
n
=∫
(
1+tan
2
x
+tanx)
n
sec
2
x
dx (as sec
2
θ=1+tan
2
θ)
Now, let t=tanx⇒dt=sec
2
xdx, we then have
I
n
=∫
(
1+t
2
+t)
n
1
dt
Making a hyperbolic substitution t=sinhy⟹dt=coshydy , we have
I
n
=∫
(
1+sinh
2
y
+sinhy)
n
coshy
dy
Now, as sinhy=
2
e
y
−e
−y
and 1+sinh
2
y=cosh
2
y, we get
I
n
=
2
1
∫
e
ny
e
y
+e
−y
dy
∴I
n
=
2
1
∫[e
−(n−1)y
+e
−(n+1)y
]dy
∴I
n
=−
2(n−1)
e
−(n−1)y
−
2(n+1)
e
−(n+1)y
+C
Now, re-substitute coshy−sinhy=e
−y
and sinhy=tanx, coshy=secx, we have
I
n
=−
2(n−1)
(secx−tanx)
(n−1)
−
2(n+1)
(secx−tanx)
(n+1)
+C
Applying the limits, we get
I
n
=[−
2(n−1)
(secx−tanx)
(n−1)
−
2(n+1)
(secx−tanx)
(n+1)
]
0
π/2
∴I
n
=
n
2
−1
n
(Note that lim
x→∞
secx−tanx=0)
Hence, proved