JEE Exam > JEE Notes > Mathematics (Maths) for JEE Main & Advanced > Important Formulas: Integration by Reduction Formula
Important Integration by Reduction Formula Formulas for JEE and NEET
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
.
Read More
|
209 videos|443 docs|143 tests
|
Related Exams
About this Document
|
4.73/5 Rating |
|
Nov 21, 2024 Last updated |
Document Description: Important Formulas: Integration by Reduction Formula for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation.
The notes and questions for Important Formulas: Integration by Reduction Formula have been prepared according to the JEE exam syllabus. Information about Important Formulas: Integration by Reduction Formula covers topics
like and Important Formulas: Integration by Reduction Formula Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Important Formulas: Integration by Reduction Formula.
Introduction of Important Formulas: Integration by Reduction Formula in English is available as part of our Mathematics (Maths) for JEE Main & Advanced
for JEE & Important Formulas: Integration by Reduction Formula in Hindi for Mathematics (Maths) for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Important Integration by Reduction Formula Formulas for JEE and NEET
Description
Full syllabus notes, lecture & questions for Important Integration by Reduction Formula Formulas for JEE and NEET - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) for JEE Main & Advanced | Best notes, free PDF download
Information about Important Formulas: Integration by Reduction Formula
In this doc you can find the meaning of Important Formulas: Integration by Reduction Formula defined & explained in the simplest way possible. Besides explaining types of
Important Formulas: Integration by Reduction Formula theory, EduRev gives you an ample number of questions to practice Important Formulas: Integration by Reduction Formula tests, examples and also practice JEE
tests
|
209 videos|443 docs|143 tests
|
Download as PDF
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Summary
,video lectures
,Previous Year Questions with Solutions
,mock tests for examination
,MCQs
,Exam
,Sample Paper
,shortcuts and tricks
,Extra Questions
,practice quizzes
,study material
,past year papers
,Viva Questions
,Important questions
,Important Integration by Reduction Formula Formulas for JEE and NEET
,Objective type Questions
,Important Integration by Reduction Formula Formulas for JEE and NEET
,Important Integration by Reduction Formula Formulas for JEE and NEET
,Semester Notes
,Free
,ppt
;
Additional Information about Important Formulas: Integration by Reduction Formula for JEE Preparation
Important Formulas: Integration by Reduction Formula Free PDF Download
The Important Formulas: Integration by Reduction Formula is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Important Formulas: Integration by Reduction Formula now and kickstart your journey towards success in the JEE exam.
Importance of Important Formulas: Integration by Reduction Formula
The importance of Important Formulas: Integration by Reduction Formula cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Important Formulas: Integration by Reduction Formula Notes
Important Formulas: Integration by Reduction Formula Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Important Formulas: Integration by Reduction Formula.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Important Formulas: Integration by Reduction Formula Notes on EduRev are your ultimate resource for success.
Important Formulas: Integration by Reduction Formula JEE Questions
The "Important Formulas: Integration by Reduction Formula JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Important Formulas: Integration by Reduction Formula on the App
Students of JEE can study Important Formulas: Integration by Reduction Formula alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Important Formulas: Integration by Reduction Formula,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Important Formulas: Integration by Reduction Formula is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup