Fo/u fopkjr Hkh# tu] ugh vkjEHk dke] foifr n[k NkM rjr e ...

fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';keA iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA

jfpr% ekuo /keZ iz.ksrk ln~xq# Jh j.kNksM+nklth egkjkt

STUDY PACKAGE Subject : Mathematics

Topic : Idefinite & Definite Integration Available Online :

R

Index 1. Theory 2. Short Revision 3. Exercise (Ex. 1 + 5 = 6) 4. Assertion & Reason 5. Que. from Compt. Exams 6. 39 Yrs. Que. from IIT-JEE(Advanced) 7. 15 Yrs. Que. from AIEEE (JEE Main)

Student's Name :______________________

Class

:______________________

Roll No.

:______________________

Address : Plot No. 27, III- Floor, Near Patidar Studio,

Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal

: 0 903 903 7779, 98930 58881, WhatsApp 9009 260 559





Get Solution of These Packages & Learn by Video Tutorials on

Indefinite Integration

FREE Download Study Package from website: &

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 2 of 89

1.

2.

(i) (iii) (v) (vii) (ix) (xi) (xii) (xiii) (xiv) (xv) (xvii) (xix)

If f & g are functions of x such that g(x) = f(x) then,

f(x) dx = g(x) + c d {g(x)+c} = f(x), where c is called the constant of integration. dx

Standard Formula:

ax bn1

(ax + b)n dx = a n 1 + c, n 1

(ii)

dx 1 = ln (ax + b) + c ax b a

1 eax+b dx = eax+b + c a

(iv)

apx+q

dx

=

1 p

apxq n a

+ c; a > 0

sin (ax + b) dx = 1 cos (ax + b) + c a

(vi)

1 cos (ax + b) dx = sin (ax + b) + c

a

tan(ax + b) dx = 1 ln sec (ax + b) + c a

(viii)

cot(ax + b) dx = 1 ln sin(ax + b)+ c

a

1 sec? (ax + b) dx = tan(ax + b) + c a

(x)

1 sec (ax + b). tan (ax + b) dx = sec (ax + b) + c a

cosec?(ax + b) dx = 1 cot(ax + b)+ c a

cosec (ax + b). cot (ax + b) dx = 1 cosec (ax + b) + c a

secx dx = ln (secx + tanx) + c

OR

ln

tan

4

2x

+

c

cosec x dx = ln (cosecx cotx) + c OR ln tan x + c OR ln (cosecx + cotx) + c 2

d x = sin1 x + c

a2 x2

a

(xvi)

dx

1

=

tan1 x

+ c

a2 x2 a

a

dx

1

=

sec1 x

+ c

x x2 a2 a

a

(xviii)

d x = ln x x2 a2 x2 a2

OR sinh1 x + c a

d x = ln x x2 a2 x2 a2

OR

cosh1 x + c

a

(xx)

d x 1 ax

a2 x2

=

2a

ln

ax

+ c

(xxi)

dx

1

xa

x2 a2

=

2a

ln

xa

+ c

(xxii)

a2 x2

dx

x

=

a2 x2

a2

+

sin1 x + c

2

2

a

(xxiii)

x2 a2 dx = x 2

x2 a2

a2

+

2

n

x

x2 a2

a

+ c

(xxiv)

x2 a2 dx = x 2

x2 a2

a2

2

n

x

x2 a2

a

+c

(xxv)

e ax eax. sin bx dx = a2 b2

(a sin bx b cos bx) + c

(xxvi)

eax eax. cos bx dx = a2 b2 (a cos bx + b sin bx) + c

3. Theorems on integration

(i)

(iii) Note : (i)

(ii)

c f(x).dx = c f(x).dx

(ii)

(f(x) g(x))dx = f(x)dx g(x)dx

f(x)dx g(x) c

f(ax b)dx

=

g(ax b) a

+ c

every contineous function is integrable

the integral of a function reffered only by a constant.

Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FREE Download Study Package from website: &

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 3 of 89

Get Solution of These Packages & Learn by Video Tutorials on

f(x).dx = g(x) + c = h(x) + c

g(x) = f(x) & g(x) ? h(x) = 0 means, g(x) ? h(x) = c

h(x) = f(x)

Example : Solution.

Evaluate : 4x5 dx

4x5 dx

=

4 6

x6 + C =

2 3

x6 + C.

Example : Solution.

Example : Solution.

Example : Solution. Example: Solution.

Evaluate :

x3

5x2

4

7 x

2 x

dx

x3

5x2

4

7 x

2 x

dx

= x3 dx + 5x2 dx ? 4dx + 7 dx + x

2 dx x

= x3 dx + 5 . x2 dx ? 4 . 1. dx + 7 . 1 dx + 2 . x1/ 2dx x

=

x4 4

+ 5 .

x3 3

?

4x

+

7

log

|

x

|

+

2

x1/ 2 1/ 2

+

C

=

x4 4

+

5 3 x3

? 4x + 7 log | x | + 4

x +C

Evaluate : ex loga ea log x ealoga dx

We have,

ex loga ea log x ealoga dx

= elogax elog xa elogaa dx

= (ax xa aa ) dx

= ax dx + xa dx + aa dx

ax

x a 1

= loga + a 1 + aa . x + C.

2x 3x

Evaluate :

5x dx

2x 3x 5x dx

=

2x 5x

3x 5x

dx

=

2

x

3

x

5 5

(2 / 5)x

(3 / 5)x

dx = loge 2 / 5 + loge 3 / 5 + C

Evaluate : sin3 x cos3 x dx

1

= 8

(2sin x cos x)3 dx

1

= 8

sin3 2x dx

1

= 32

(3 sin2x sin 6x) dx

1 3 sin2x sin6x

= 8

4

dx

1 = 32

3 2

cos

2x

1 6

cos

6x

+

C

Example :

x4

Evaluate : x2 1 dx

Solution.

x4 x2 1 dx

=

x4 11 x2 1 dx =

x4 1

1

x2 1 + x2 1 dx =

(x2 1) dx +

1

x3

x2 1 dx = 3 ? x + tan?1 x + C

Example: Solution.

1

Evaluate : 4 9x2 dx

We have

1 4 9x2

Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FREE Download Study Package from website: &

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 4 of 89

Get Solution of These Packages & Learn by Video Tutorials on

1

1

= 9 4 x2 dx

9

1

1

= 9 (2 / 3)2 x2 dx

=

1 9

.

1 (2 / 3)

tan?1

x 2/3

+ C =

1 6

tan?1

3x 2

+ C

Example :

cos x cos 2x dx

Solution.

cos x cos2x dx

1

=

2cos x cos 2x dx

2

1

=

(cos3x cos x) dx

2

Self Practice Problems

=

1 2

sin3x sin x

3

1

+ c

1.

Evaluate : tan2 x dx

Ans. tanx ? x + C

2.

1

Evaluate : 1 sin x dx

Ans. tanx ? sec x + C

4. Integration by Subsitutions

If we subsitute x = (t) in a integral then

(i) everywhere x will be replaced in terms of t. (ii) dx also gets converted in terms of dt.

(iii) (t) should be able to take all possible value that x can take.

Example : Solution.

Evaluate : x3 sin x4 dx

We have

= x3 sin x4 dx

Example : Solution.

Let x4 = t

(n x)2 dx x

(n x)2 dx x

Put nx = t

=

t2. dx x

t3 = +c

3

d(x4) = dt

1

4x3 dx = dt dx = 4x3 dt

1

dx = dt

x

= t2dt

(n x)3

=

+ c

3

Example : Solution.

Example : Solution.

Evaluate (1 sin2 x)cos x dx

Put sinx = t cosx dx = dt

(1 t2 ) dt = t + t3 + c 3

sin3 x

= sin x +

+ c

3

x

Evaluate : x4 x2 1 dx We have,

x

x

= x 4 x2 1 dx = (x2 )2 x2 1 dx

dt

Let x2 = t, then, d (x2) = dt

2x dx = dt

dx =

2x

x

dt

= t2 t 1 . 2x

1

1

= 2

t2 t 1 dt

=

1 2

1

t

1 2

2

3 2 2

dt

1 = 2.

1 3

tan?1

t 1

2 3

+

C

2

2

Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FREE Download Study Package from website: &

Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. page 5 of 89

Get Solution of These Packages & Learn by Video Tutorials on

=

1 3

tan?1

2 t

3

1

+

C =

1 3

tan?1

2x2 1 3

+ C.

Note: (i) (iii) (iv)

[ f(x)]n f (x) dx = (f(x))n1 n1

(ii)

f (x)

f (x)n

d x n N Take xn common & put 1 + xn = t.

x (xn 1)

dx

(n1)

n N, take xn common & put 1+xn = tn

x2 xn 1 n

( f ( x))1n dx =

1 n

(v)

xn

dx 1 xn 1/ n

take xn common as x and put 1 + xn = t.

Self Practice Problems

1.

sec 2 x dx 1 tan x

Ans. n |1 + tan x| + C

2.

sin(nx) dx x

Ans. ? cos (n x) + C

5. Integration by Part :

f(x) g(x) dx = f(x)

g(x) dx ?

d f(x) g(x) dx dx

dx

(i)

when you find integral g(x)dx then it will not contain arbitarary constant.

(ii)

g(x)dx should be taken as same both terms.

(iii) the choice of f(x) and g(x) is decided by ILATE rule.

the function will come later is taken an integral function.

Inverse function

L

Logrithimic function

A

Algeberic function

T

Trigonometric function

E

Exponential function

Example : Solution.

Example : Solution.

Evaluate : x tan1 x dx

x tan1 x dx

= (tan?1 x) x2 ? 2

1

x2

1 x2 . 2 dx

x2

1

=

tan?1 x ?

2

2

x2 11

x2

1

x2 1

dx = 2

tan?1 x ? 2

1

1 x2 1

dx

x2

1

=

tan?1 x ? [x ? tan?1 x] + C.

2

2

Evaluate : x log(1 x) dx

x log(1 x) dx

x2

1 x2

= log (x + 1) . ?

. dx

2

x1 2

x2

1

= log (x + 1) ?

x2

x2

1

dx = log (x + 1) ?

x2 1 1 dx

2

2 x1

2

2 x 1

x2

1 x2 1

1

= log (x + 1) ?

+

dx

2

2 x 1 x1

=

x2 2

log (x + 1) ?

1 2

( x

1)

x

1

1

dx

=

x2 2

log (x + 1) ?

1 2

x2

2

x log | x 1|

+ C

Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download