⇒
∫ (n − 1) sin
n−2
Ιn = sinn–1 x . (– cos x) +
⇒ ⇒
= – cos x sinn–1 x + (n - 1)
∫ sin
= – cos x sinn–1 x + (n – 1)
∫ sin
n−2
x cos 2 x dx
x(1 − sin2 x ) dx
n− 2
∫
x dx − (n − 1) sinn x dx
Ιn = – cos x sin x + (n – 1) Ιn–2 – (n – 1) Ιn nΙn = – cos x . sinn–1 x + (n – 1) Ιn–2 ; n–1
Ιn = –
n −1 cos x sinn−1 x + Ιn–2 n n
Example : 44 Find the reduction formula for If Ιn =
∫ tan
n
∫ tan
n
x dx .
x dx , to prove that (n – 1) (Ιn + Ιn–2) = tann–1 x.
Solution Here Ιn =
∫ tan
=
∫ tan
x(sec 2 x − 1) dx
=
∫ tan
x sec 2 x dx –
=
∫ tan
n
x dx
n −2
n− 2
n− 2
∫ tan
n− 2
x tan2 x dx
∫ tan
n −2
x dx
x sec 2 x − Ι n− 2
tann−1 x n −1 Hence (n – 1) (Ιn + Ιn–2) = tann–1 x. ⇒
Ιn + Ιn–2 =
Example : 45
∫ sec
Find reduction formula for
n
x dx
Solution Let Ιn =
∫ sec
⇒
Ιn =
n
x
∫ sec
n−2
x sec2x dx
Apply by parts taking secn–2 x as the first part and sec2x as the second part
∫ sec
3
⎡d
∫ ⎢⎣ dx (sec
n−2
⇒
Ιn = secn–2 x
⇒
Ιn = secn–2 x tan x –
⇒
Ιn = secn–2 x tan x – (n – 2)
⇒
Ιn + (n – 2) Ιn = secn–2 x tan x + (n – 2)
⇒
(n – 1) Ιn = secn–2 x tan x + (n – 2) Ιn–2
x dx –
∫ (n − 2) sec
n− 3
∫ sec
⎤ x ) sec 2 x dx ⎥ dx ⎦
∫
x sec x tan x tan x dx
n −2
x (sec 2 x − 1) dx
∫ sec
n− 2
x dx
Page # 23.