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Important Integration by Reduction Formula Formulas for JEE and NEET

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 Page 1


 
 
F. INTEGRATION BY REDUCTION FORMULAE
Ex.47 If I
n
 = 
?
?
2 2 n
x a x dx, prove that I
n
 = –
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
+
) 2 n (
) 1 n (
?
?
 a
2
 I
n–2
.
Sol. I
n
 = 
?
?
2 2 n
x a x dx = 
?
?
?
dx } x a x .{ x
2 2 1 n
Applying integration by parts we get
= x
n–1
 . 
2 2 3/2
(a x )
3
? ?
? ? ?
? ?
?
? ?
? ?
 +
n 2
(n 1)x
?
?
?
. 
2 2 3/2
(a x )
3
? ?
? ? ?
?
? ?
? ?
? ?
 dx
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
) 1 n ( ?
?
?
?
) x a .( x
2 2 2 n
2 2
x a ?
dx
? I
n 
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
 – 
3
) 1 n ( ?
 I
n
? I
n
 + 
3
) 1 n ( ?
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
a ) 1 n (
2
?
 I
n–2
?
?
?
?
?
?
? ?
3
2 n
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
? I
n
 = 
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
 + 
) 2 n (
a ) 1 n (
2
?
?
 I
n–2
 
Page 2


 
 
F. INTEGRATION BY REDUCTION FORMULAE
Ex.47 If I
n
 = 
?
?
2 2 n
x a x dx, prove that I
n
 = –
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
+
) 2 n (
) 1 n (
?
?
 a
2
 I
n–2
.
Sol. I
n
 = 
?
?
2 2 n
x a x dx = 
?
?
?
dx } x a x .{ x
2 2 1 n
Applying integration by parts we get
= x
n–1
 . 
2 2 3/2
(a x )
3
? ?
? ? ?
? ?
?
? ?
? ?
 +
n 2
(n 1)x
?
?
?
. 
2 2 3/2
(a x )
3
? ?
? ? ?
?
? ?
? ?
? ?
 dx
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
) 1 n ( ?
?
?
?
) x a .( x
2 2 2 n
2 2
x a ?
dx
? I
n 
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
 – 
3
) 1 n ( ?
 I
n
? I
n
 + 
3
) 1 n ( ?
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
a ) 1 n (
2
?
 I
n–2
?
?
?
?
?
?
? ?
3
2 n
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
? I
n
 = 
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
 + 
) 2 n (
a ) 1 n (
2
?
?
 I
n–2
 
INDEFINITE INTEGRATION
Ex.48 Integration of 1/(x
2
 + k)
n
.
Sol. Thus 
? ?
?
1 n 2
) k x (
1
 1 . dx 
1 n 2
) k x (
x
?
?
 – 
2 n
(n 1)
x.
(x k)
? ?
?
?
 . 2x dx
or I
n–1
 = 
1 n 2
) k x (
x
?
?
+ 2(n – 1) 
?
?
? ?
n 3
2
) k x (
k ) k x (
dx, [ ?  x
2
 = (x
2
 + k) – k]
or I
n–1
 = 
1 n 2
) k x (
x
?
?
 + 2 (n – 1) 
?
?
?
?
?
?
?
?
?
?
?
? ? ? n 2 1 n 2
) k x (
dx
k
) k x (
dx
or I
n–1
 = 
1 n 2
) k x (
x
?
?
 + 2 (n – 1) l
n–1
 – 2k (n –1) l
n
. ? ? ? ? ? 2k (n–1) l
n
 =
1 n 2
) k x (
x
?
?
+ {2(n–1) – 1} l
n–1
or 2k(n – 1) l
n
 = 
1 n 2
) k x (
x
?
?
+ (2n – 3)
 
I
n–1
.
Hence 
? ?
?
1 n 2
) k x (
dx
 = 
1 n 2
) k x )( 1 n ( k 2
x
?
? ?
 + 
) 1 n ( k 2
) 3 n 2 (
?
?
? ?
?
1 n 2
) k x (
dx
.
Above is the reduction formula for 
?
? dx ] ) k x /( 1 [
n 2
. By repeated application of this formula the
integral shall reduce to that of 
) k x (
1
2
?
 which is 
k
1
 tan
–1
 ?
?
?
?
?
?
?
?
k
x
.
Ex.49 Integrate (x + 2) / (2x
2
 + 4x + 3)
2
.
Sol. Here (d/dx) (2x
2
 + 4x + 3) = 4x + 4.
?
?
? ?
?
2 2
) 3 x 4 x 2 (
dx ) 2 x (
= 
?
? ?
? ? ?
2 2
) 3 x 4 x ( 2
1 2 ) 4 x 4 (
4
1
dx = 
4
1
?
? ?
?
2 2
) 3 x 4 x 2 (
dx ) 4 4 (
= 
?
?
?
?
?
?
?
? ?
?
2
2
2
3
x 2 x
dx ) 1 2 (
= 
4
1
?
?
? ?
2 2
) 3 x 4 x 2 (
(4x + 4) dx + 
4
1
?
?
?
?
?
?
?
? ?
2
2
2
3
x 2 x
dx
= – 
) 3 x 4 x 2 ( 4
1
2
? ?
+
4
1
?
?
?
?
?
?
?
? ?
2
2
2
1
) 1 x (
dx
Now put x + 1 = t and then applying the reduction formula, we get
I = 
) 3 x 4 x 2 ( 4
1
2
? ?
 + 
4
1 c } ) 1 x ( 2 { tan 2
2
1
) 1 x (
) 1 x (
1
2
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
Ex.50 Integrate (2x + 3)/(x
2
 + 2x + 3)
2
.
Sol. Here (d/dx) (x
2
 + 2x + 3) = 2x + 2
? I = ?
? ?
?
2 2
) 3 x 2 x (
) 3 x 2 (
 = 
2 2
(2x 2 1)dx
(x 2x 3)
? ?
? ?
? = ?
? ?
?
2 2
) 3 x 2 x (
dx ) 2 x 2 (
= ?
? ?
2 2
) 3 x 2 x (
dx
= – 
) 3 x 2 x (
1
2
? ?
 + ?
? ?
2 2
) 3 x 2 x (
dx
....(i)
Now let I
1
 = 
?
? ?
2 2 2
] 2 ) 1 x [(
dx
 (Put x + 1 = 
2
 tan t, so that dx = 
2
 sec
2
 t dt)
 
Page 3


 
 
F. INTEGRATION BY REDUCTION FORMULAE
Ex.47 If I
n
 = 
?
?
2 2 n
x a x dx, prove that I
n
 = –
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
+
) 2 n (
) 1 n (
?
?
 a
2
 I
n–2
.
Sol. I
n
 = 
?
?
2 2 n
x a x dx = 
?
?
?
dx } x a x .{ x
2 2 1 n
Applying integration by parts we get
= x
n–1
 . 
2 2 3/2
(a x )
3
? ?
? ? ?
? ?
?
? ?
? ?
 +
n 2
(n 1)x
?
?
?
. 
2 2 3/2
(a x )
3
? ?
? ? ?
?
? ?
? ?
? ?
 dx
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
) 1 n ( ?
?
?
?
) x a .( x
2 2 2 n
2 2
x a ?
dx
? I
n 
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
 – 
3
) 1 n ( ?
 I
n
? I
n
 + 
3
) 1 n ( ?
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
a ) 1 n (
2
?
 I
n–2
?
?
?
?
?
?
? ?
3
2 n
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
? I
n
 = 
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
 + 
) 2 n (
a ) 1 n (
2
?
?
 I
n–2
 
INDEFINITE INTEGRATION
Ex.48 Integration of 1/(x
2
 + k)
n
.
Sol. Thus 
? ?
?
1 n 2
) k x (
1
 1 . dx 
1 n 2
) k x (
x
?
?
 – 
2 n
(n 1)
x.
(x k)
? ?
?
?
 . 2x dx
or I
n–1
 = 
1 n 2
) k x (
x
?
?
+ 2(n – 1) 
?
?
? ?
n 3
2
) k x (
k ) k x (
dx, [ ?  x
2
 = (x
2
 + k) – k]
or I
n–1
 = 
1 n 2
) k x (
x
?
?
 + 2 (n – 1) 
?
?
?
?
?
?
?
?
?
?
?
? ? ? n 2 1 n 2
) k x (
dx
k
) k x (
dx
or I
n–1
 = 
1 n 2
) k x (
x
?
?
 + 2 (n – 1) l
n–1
 – 2k (n –1) l
n
. ? ? ? ? ? 2k (n–1) l
n
 =
1 n 2
) k x (
x
?
?
+ {2(n–1) – 1} l
n–1
or 2k(n – 1) l
n
 = 
1 n 2
) k x (
x
?
?
+ (2n – 3)
 
I
n–1
.
Hence 
? ?
?
1 n 2
) k x (
dx
 = 
1 n 2
) k x )( 1 n ( k 2
x
?
? ?
 + 
) 1 n ( k 2
) 3 n 2 (
?
?
? ?
?
1 n 2
) k x (
dx
.
Above is the reduction formula for 
?
? dx ] ) k x /( 1 [
n 2
. By repeated application of this formula the
integral shall reduce to that of 
) k x (
1
2
?
 which is 
k
1
 tan
–1
 ?
?
?
?
?
?
?
?
k
x
.
Ex.49 Integrate (x + 2) / (2x
2
 + 4x + 3)
2
.
Sol. Here (d/dx) (2x
2
 + 4x + 3) = 4x + 4.
?
?
? ?
?
2 2
) 3 x 4 x 2 (
dx ) 2 x (
= 
?
? ?
? ? ?
2 2
) 3 x 4 x ( 2
1 2 ) 4 x 4 (
4
1
dx = 
4
1
?
? ?
?
2 2
) 3 x 4 x 2 (
dx ) 4 4 (
= 
?
?
?
?
?
?
?
? ?
?
2
2
2
3
x 2 x
dx ) 1 2 (
= 
4
1
?
?
? ?
2 2
) 3 x 4 x 2 (
(4x + 4) dx + 
4
1
?
?
?
?
?
?
?
? ?
2
2
2
3
x 2 x
dx
= – 
) 3 x 4 x 2 ( 4
1
2
? ?
+
4
1
?
?
?
?
?
?
?
? ?
2
2
2
1
) 1 x (
dx
Now put x + 1 = t and then applying the reduction formula, we get
I = 
) 3 x 4 x 2 ( 4
1
2
? ?
 + 
4
1 c } ) 1 x ( 2 { tan 2
2
1
) 1 x (
) 1 x (
1
2
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
Ex.50 Integrate (2x + 3)/(x
2
 + 2x + 3)
2
.
Sol. Here (d/dx) (x
2
 + 2x + 3) = 2x + 2
? I = ?
? ?
?
2 2
) 3 x 2 x (
) 3 x 2 (
 = 
2 2
(2x 2 1)dx
(x 2x 3)
? ?
? ?
? = ?
? ?
?
2 2
) 3 x 2 x (
dx ) 2 x 2 (
= ?
? ?
2 2
) 3 x 2 x (
dx
= – 
) 3 x 2 x (
1
2
? ?
 + ?
? ?
2 2
) 3 x 2 x (
dx
....(i)
Now let I
1
 = 
?
? ?
2 2 2
] 2 ) 1 x [(
dx
 (Put x + 1 = 
2
 tan t, so that dx = 
2
 sec
2
 t dt)
 
INDEFINITE INTEGRATION
? I
1
 = 
?
?
2 2
2
) 2 t tan 2 (
dt t sec 2
 = 
4
2
?
dt t cos
2
 = 
4
2
 
?
2
1
 (1 + cos 2t) dt
= 
8
2
 [+ 
2
1
 sin 2t] + 
8
2
 [t + sin t cos t] + c
Now tan t = 
2
1 x ?
. Therefore sin t = 
} 2 ) 1 x {(
1 x
2
? ?
?
= 
) 3 x 2 x (
1 x
2
? ?
?
 , and cos t = 
) 3 x 2 x (
2
2
? ?
Also t = tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
. Hence I
1
 = 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
 + 
8
2
. 
)} 3 x 2 x {(
1 x
2
? ?
?
. 
) 3 x 2 x (
2
2
? ?
.
= 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
 + 
4
1
 
) 3 x 2 x (
1 x
2
? ?
?
? I = – 
3 x 2 x
1
2
? ?
 + 
4
1
3 x 2 x
1 x
2
? ?
?
 + 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
+ c, from (i)
= 
) 3 x 2 x ( 4
4 1 x
2
? ?
? ?
 + 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
+ c = 
) 3 x 2 x ( 4
3 x
2
? ?
?
 + 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
+ c
Ex.51 If I
m
 = 
?
?
m
) x cos x (sin
 dx, then show that mI
m
 = (sinx + cosx)
m–1
. (sinx – cosx) + 2 (m – 1) I
m–2
Sol. ? I
m
 = 
?
?
m
) x cos x (sin
dx
= 
?
?
?
1 m
) x cos x (sin
. (sinx + cosx) dx, applying integration by parts.
= (sin x + cosx)
m–1
 (cosx + sin x) – 
m 2
(m 1)(sinx cosx) dx
?
? ?
?
 . (cosx – sin x) . (sinx – cosx) dx
= (sinx + cosx)
m–1
 (sinx – cos x) + (m – 1) 
?
?
?
2 m
) x cos x (sin
(sinx + cos x)
2
 . dx
As we know, (sinx + cosx)
2
 + (sinx – cosx)
2
 = 2,
? I
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + (m – 1) 
?
?
?
2 m
) x cos x (sin
 {2 – (sinx + cosx)
2
}
 
dx
= (sinx + cosx)
m–1
 (sin x – cos x) + (m – 1) 
?
?
?
2 m
) x cos x (sin 2
 dx – (m – 1) 
?
? dx ) x cos x (sin
m
I
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + (m – 1)I
m–2
 – (m – 1)I
m
or (m – 1) I
m
 + I
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + 2 (m – 1) I
m–2
or mI
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + 2 (m – 1) I
m–2
Ex.52 If I
m,n
 = 
?
, dx . nx cos . x cos
m
 show that (m + n) I
m,n
 = cos
m
x. sin nx + m I
(m–1, n–1)
Sol. We have, I
m,n
 = 
m
cos x.cosnx
?
 dx
= (cos
m
x) ?
?
?
?
?
?
n
nx sin
?
?1 m
cos m
(–sinx).
n
nx sin
 dx = 
n
1
 cos
m
 x . sinnx + 
n
m
m 1
cos x(sinx.sinnx)dx
?
?
As we have cos (n – 1) x = cos nx cos x + sin nx . sinx
 
Page 4


 
 
F. INTEGRATION BY REDUCTION FORMULAE
Ex.47 If I
n
 = 
?
?
2 2 n
x a x dx, prove that I
n
 = –
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
+
) 2 n (
) 1 n (
?
?
 a
2
 I
n–2
.
Sol. I
n
 = 
?
?
2 2 n
x a x dx = 
?
?
?
dx } x a x .{ x
2 2 1 n
Applying integration by parts we get
= x
n–1
 . 
2 2 3/2
(a x )
3
? ?
? ? ?
? ?
?
? ?
? ?
 +
n 2
(n 1)x
?
?
?
. 
2 2 3/2
(a x )
3
? ?
? ? ?
?
? ?
? ?
? ?
 dx
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
) 1 n ( ?
?
?
?
) x a .( x
2 2 2 n
2 2
x a ?
dx
? I
n 
= – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
 – 
3
) 1 n ( ?
 I
n
? I
n
 + 
3
) 1 n ( ?
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
+
3
a ) 1 n (
2
?
 I
n–2
?
?
?
?
?
?
? ?
3
2 n
I
n
 = – 
3
) x a ( x
2 / 3 2 2 1 n
?
?
 +
3
a ) 1 n (
2
?
 I
n–2
? I
n
 = 
) 2 n (
) x a ( x
2 / 3 2 2 1 n
?
?
?
 + 
) 2 n (
a ) 1 n (
2
?
?
 I
n–2
 
INDEFINITE INTEGRATION
Ex.48 Integration of 1/(x
2
 + k)
n
.
Sol. Thus 
? ?
?
1 n 2
) k x (
1
 1 . dx 
1 n 2
) k x (
x
?
?
 – 
2 n
(n 1)
x.
(x k)
? ?
?
?
 . 2x dx
or I
n–1
 = 
1 n 2
) k x (
x
?
?
+ 2(n – 1) 
?
?
? ?
n 3
2
) k x (
k ) k x (
dx, [ ?  x
2
 = (x
2
 + k) – k]
or I
n–1
 = 
1 n 2
) k x (
x
?
?
 + 2 (n – 1) 
?
?
?
?
?
?
?
?
?
?
?
? ? ? n 2 1 n 2
) k x (
dx
k
) k x (
dx
or I
n–1
 = 
1 n 2
) k x (
x
?
?
 + 2 (n – 1) l
n–1
 – 2k (n –1) l
n
. ? ? ? ? ? 2k (n–1) l
n
 =
1 n 2
) k x (
x
?
?
+ {2(n–1) – 1} l
n–1
or 2k(n – 1) l
n
 = 
1 n 2
) k x (
x
?
?
+ (2n – 3)
 
I
n–1
.
Hence 
? ?
?
1 n 2
) k x (
dx
 = 
1 n 2
) k x )( 1 n ( k 2
x
?
? ?
 + 
) 1 n ( k 2
) 3 n 2 (
?
?
? ?
?
1 n 2
) k x (
dx
.
Above is the reduction formula for 
?
? dx ] ) k x /( 1 [
n 2
. By repeated application of this formula the
integral shall reduce to that of 
) k x (
1
2
?
 which is 
k
1
 tan
–1
 ?
?
?
?
?
?
?
?
k
x
.
Ex.49 Integrate (x + 2) / (2x
2
 + 4x + 3)
2
.
Sol. Here (d/dx) (2x
2
 + 4x + 3) = 4x + 4.
?
?
? ?
?
2 2
) 3 x 4 x 2 (
dx ) 2 x (
= 
?
? ?
? ? ?
2 2
) 3 x 4 x ( 2
1 2 ) 4 x 4 (
4
1
dx = 
4
1
?
? ?
?
2 2
) 3 x 4 x 2 (
dx ) 4 4 (
= 
?
?
?
?
?
?
?
? ?
?
2
2
2
3
x 2 x
dx ) 1 2 (
= 
4
1
?
?
? ?
2 2
) 3 x 4 x 2 (
(4x + 4) dx + 
4
1
?
?
?
?
?
?
?
? ?
2
2
2
3
x 2 x
dx
= – 
) 3 x 4 x 2 ( 4
1
2
? ?
+
4
1
?
?
?
?
?
?
?
? ?
2
2
2
1
) 1 x (
dx
Now put x + 1 = t and then applying the reduction formula, we get
I = 
) 3 x 4 x 2 ( 4
1
2
? ?
 + 
4
1 c } ) 1 x ( 2 { tan 2
2
1
) 1 x (
) 1 x (
1
2
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
Ex.50 Integrate (2x + 3)/(x
2
 + 2x + 3)
2
.
Sol. Here (d/dx) (x
2
 + 2x + 3) = 2x + 2
? I = ?
? ?
?
2 2
) 3 x 2 x (
) 3 x 2 (
 = 
2 2
(2x 2 1)dx
(x 2x 3)
? ?
? ?
? = ?
? ?
?
2 2
) 3 x 2 x (
dx ) 2 x 2 (
= ?
? ?
2 2
) 3 x 2 x (
dx
= – 
) 3 x 2 x (
1
2
? ?
 + ?
? ?
2 2
) 3 x 2 x (
dx
....(i)
Now let I
1
 = 
?
? ?
2 2 2
] 2 ) 1 x [(
dx
 (Put x + 1 = 
2
 tan t, so that dx = 
2
 sec
2
 t dt)
 
INDEFINITE INTEGRATION
? I
1
 = 
?
?
2 2
2
) 2 t tan 2 (
dt t sec 2
 = 
4
2
?
dt t cos
2
 = 
4
2
 
?
2
1
 (1 + cos 2t) dt
= 
8
2
 [+ 
2
1
 sin 2t] + 
8
2
 [t + sin t cos t] + c
Now tan t = 
2
1 x ?
. Therefore sin t = 
} 2 ) 1 x {(
1 x
2
? ?
?
= 
) 3 x 2 x (
1 x
2
? ?
?
 , and cos t = 
) 3 x 2 x (
2
2
? ?
Also t = tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
. Hence I
1
 = 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
 + 
8
2
. 
)} 3 x 2 x {(
1 x
2
? ?
?
. 
) 3 x 2 x (
2
2
? ?
.
= 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
 + 
4
1
 
) 3 x 2 x (
1 x
2
? ?
?
? I = – 
3 x 2 x
1
2
? ?
 + 
4
1
3 x 2 x
1 x
2
? ?
?
 + 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
+ c, from (i)
= 
) 3 x 2 x ( 4
4 1 x
2
? ?
? ?
 + 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
+ c = 
) 3 x 2 x ( 4
3 x
2
? ?
?
 + 
8
2
 tan
–1
 ?
?
?
?
?
?
?
? ?
2
1 x
+ c
Ex.51 If I
m
 = 
?
?
m
) x cos x (sin
 dx, then show that mI
m
 = (sinx + cosx)
m–1
. (sinx – cosx) + 2 (m – 1) I
m–2
Sol. ? I
m
 = 
?
?
m
) x cos x (sin
dx
= 
?
?
?
1 m
) x cos x (sin
. (sinx + cosx) dx, applying integration by parts.
= (sin x + cosx)
m–1
 (cosx + sin x) – 
m 2
(m 1)(sinx cosx) dx
?
? ?
?
 . (cosx – sin x) . (sinx – cosx) dx
= (sinx + cosx)
m–1
 (sinx – cos x) + (m – 1) 
?
?
?
2 m
) x cos x (sin
(sinx + cos x)
2
 . dx
As we know, (sinx + cosx)
2
 + (sinx – cosx)
2
 = 2,
? I
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + (m – 1) 
?
?
?
2 m
) x cos x (sin
 {2 – (sinx + cosx)
2
}
 
dx
= (sinx + cosx)
m–1
 (sin x – cos x) + (m – 1) 
?
?
?
2 m
) x cos x (sin 2
 dx – (m – 1) 
?
? dx ) x cos x (sin
m
I
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + (m – 1)I
m–2
 – (m – 1)I
m
or (m – 1) I
m
 + I
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + 2 (m – 1) I
m–2
or mI
m
 = (sinx + cosx)
m–1
 (sinx – cosx) + 2 (m – 1) I
m–2
Ex.52 If I
m,n
 = 
?
, dx . nx cos . x cos
m
 show that (m + n) I
m,n
 = cos
m
x. sin nx + m I
(m–1, n–1)
Sol. We have, I
m,n
 = 
m
cos x.cosnx
?
 dx
= (cos
m
x) ?
?
?
?
?
?
n
nx sin
?
?1 m
cos m
(–sinx).
n
nx sin
 dx = 
n
1
 cos
m
 x . sinnx + 
n
m
m 1
cos x(sinx.sinnx)dx
?
?
As we have cos (n – 1) x = cos nx cos x + sin nx . sinx
 
INDEFINITE INTEGRATION
? I
m,n
 = 
n
1
 cos
m
 x . sin x + 
n
m
?
?
x cos
1 m
. {cos(n–1) x – cosnx . cosx} dx
= 
n
1
 cos
m
 x . sin x + 
n
m
?
?
?
dx . x ) 1 n cos( . x cos
1 m
 – 
n
m
?
nxdx cos . x cos
m
= 
n
1
 cos
m
 x . sin nx + 
n
m
 I
m–1,n–1
 – 
n
m
 I
m,n
? I
m,n
 + 
n
m
I
m,n
 = 
n
1
 [cos
m
x . sinnx + mI
m–1, n–1
]  ? ?
m n
n
? ? ?
? ?
? ?
I
m,n
 = 
n
1
 [cos
m
 x . sin nx + mI
m–1,n–1
]
? (m + n) I
m,n
 = cos
m
 x . sin nx + mI
m–1,n–1
.
 
 
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