9_Integrals.pdf

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INTEGRALS

5.1 BASIC CONCEPTS

If f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of f(x). i.e.,

dx

d _{{g(x)} = f(x) Û}

ò

f(x)dx = g(x).

(a) Derivative of a function is unique but a function can have infinite antiderivatives or integrals.

(b)

ò

f(x)dx=g(x)+c, _{where c is constant of integration, is knwon as indefinite integral.}

(c)

ò

b a dx ) x ( f = {g(x) + c b a } = g(b) – g(a). (d)

ò

dx=x+c. (e)

ò

c.f(x)dx=c.

ò

f(x)dx. (f)

ò

{(x)±g(x)}dx =

ò

f(x)dx ±

ò

g(x)dx. (g)

ò

f(x)dx Û

ò

f{g(t)}. g’(t)dt, if we substitute x = g(t) such that dx = g’(t)dt.

5.2 STANDARD RESULT 1.

ò

xn _{dx =} 1 n xn1 + +

+ c, n ¹ – 1, n is a rational number. If n = –1, then

ò

x 1 dx = log|x| + c. 2.

ò

(ax + b)n _{dx =}

(

)

(

n 1

)

a b ax n1 + + + + c, n ¹ – 1, n is a rational number.. If n = – 1. Then

ò

b ax 1 + dx = a 1_{log|ax + b| + c.}

3. In case of rational function, if degree of numerator is equal or greater than degree of denomina-tor, then we first divide numerator by denominator and write it as

Numerator

Denominator = Quotient + Denominator mainder

Re _{, and then integrate.}

4.

ò

sin ax dx = a ax cos + c. If a = 1,

ò

sinx dx = – cosx + c. 5.

ò

cos ax dx = a ax sin _{+ c. If a = 1,}

ò

cos x dx = sin x + c. 6.

ò

tan ax dx = – a 1 log | cos as | + c. or a 1 log | sec ax | + c. If a = 1, then

ò

tan x dx = – log | cos x | + c or log | sec x | + c.

7.

ò

cot ax dx =

a

1_{log | sin ax | + c. If a = 1, then}

ò

cot x dx = log | sin x | + c.

8.

ò

sec ax dx =

a 1

log | sec ax + tan ax | + c. If a = 1, then

ò

sec x dx = log | sec x + tan x | + c.

9.

ò

cosec ax dx =

a

1_{log | cosec ax – cot ax | + c. If a = 1, then}

ò

cosec x dx = log | cosec x – cot x |+c.

10.

ò

sec ax tan ax dx =

a

1_{sec ax + c. If a = 1, then}

ò

sec x tan x dx = sec x + c.

11.

ò

cosec ax cot ax dx = –

a

1_{cosec ax + c. If a = 1, then}

ò

cosec x cot x dx = –cosec x + c.

12.

ò

sec2 _{ax dx =} a 1_{tan ax + c. If a = 1, then}

ò

sec2 _{x dx = tan x + c.} 13.

ò

cosec2 _{ax dx = –} a 1

cot ax + c. If a = 1, then

ò

cosec2 _{x dx = –cot x + c.}

14.

ò

eax _{dx =}

a eax

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15.

ò

amx _{dx =} a log m a e mx + c. If m = 1, then

ò

ax _{dx =} a log a e x + c. 16.

ò

_a2 _x2 1 - dx = sin–1 a x + c. If a = 1, then

ò

_{1 x}2 1 - dx = sin–1x + c. 17.

ò

2 2 x a 1 + dx = a 1_tan–1 a x _{+ c. If a = 1, then}

ò

_{1 x}2 1 + dx = tan–1x + c. 18.

ò

_{x x}2 _a2 1 - dx = a 1 sec–1 a x + c. If a = 1, then

ò

1 x x 1 2_- dx = sec–1x + c. 19.

ò

2 2 1 a +x dx = log 2 2 _x a x+ + _{+ c. 20.}

ò

_{2 2} a x 1 - dx= log 2 2 x+ x -a _{+ c.} 21.

ò

_x2 _a2 1 - dx = x a a x log a 2 1 + + c. 22.

ò

_a2 _x2 1 - dx = a x x a log a 2 1 -+ + c. 23.

ò

_a2_-x2 dx = 2 x 2 2 _x a - + a₂ 2 sin–1 a x + c. 24.

ò

_a2_+x2 dx = 2 x 2 2 _x a + + a₂ 2 log 2 2 x a x+ + _{+ c.} 25.

ò

_x2_-a2 dx = 2 x 2 2 _a x - – a₂ 2 log_{x+ x}2_-a2 _{+ c.} 5.3 METHODS OF INTEGRATION

(i) Integration by Substitution (or change of independent variable) :

If the independent variable x in

ò

f(x) dx be changed to t, then we substitute x = f (t) i.e., dx = f‘(t) dt

ò

f(x) dx =

ò

f(f(t))f’(t) dt

which is either a standard form or is easier to integrate.

(ii) Integration by parts :

If u and v are the differentiable functions of x then

ò

u. v dx = u

ò

vdx –

ò

( )

ú û ù ê ë é ÷ ø ö ç è æ

ò

vdx u dx d dx. How to choose Ist and IInd functions:

(a) If the two functions are of different types take that function as Ist which comes first in the word

ILATE where I stands for inverse circular function, L stands for logarithrmic function A stands for

Algebratic function, T stands for trigonometric function and E stands for exponential function. (b) If both functions are algebratic take that function as Ist whose differential coefficient is simple. (c) It both functions are trigonometrical take that function as IInd whose integral is simpler. (d) Successive integration by parts : Use the following formula

ò

uvdx = uv₁ – u’v₂ + u”v₃ – u’” v₄ + ... ...+ (–1)n–1 un –1 v_n + (–1)n

ò

un v_ndx.

where un stands for nth differential coefficient of u w. r. t. x and v_n stands for n th integral of v w. r. t. x. cancellation of Integrals : i.e.

ò

ex{(f(x) + f’(x)} dx = exf(x) + c.

(iii) Evalution of Integrals of various Types : form the of Integrals : I Type (i)

ò

c bx ax dx 2_{+ +} (ii)

ò

(

_{ax bx c}

)

dx 2_{+ +} (iii)

ò

(

ax2+bx+c

)

dx

Rule : Express ax2 + bx + c in the form of perfect square and then apply the standard resuts.

Integrals of the form

Type II : (i)

ò

c bx ax q px 2_{+ +} + (ii)

ò

c bx ax q px 2_{+ +} + (iii)

ò

(

p x p

₍

x_ax2 ..._{bx c}p

₎

n1x pn

)

1 n 1 n 0 + + + + + + - -dx

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Rule for (I) :

ò

(

)

c bx ax dx q px 2_{+ +} + =

_{( )}

₂b_a

ò

₍

_ax

(

2ax2_+bxb

)

dx_+c

₎

+ + ÷ ø ö ç è æ -a 2 pb q

ò

c bx ax dx 2_{+ +} = a 2 p _{ln | ax}2 _{+ bx + c | + ÷} ø ö ç è æ -a 2 bp q

ò

c bx ax dx 2_{+ +} The other integral of R. H. S. can be evaluated with the help of type I.

Rule for (II) :

ò

(

)

2 px q dx ax bx c + + + = 2a p

ò

₍

(

2

)

₎

2ax b dx ax bx c + + + dx + ÷ ø ö ç è æ -a 2 pb q

ò

c bx ax dx 2_{+ +} = a p

₍

₎

c bx ax2_{+ + +} ÷ ø ö ç è æ -a 2 bp q

ò

c bx ax dx 2_{+ +}

The other integral of R. H. S. can be evaluated with the help of type I.

Rule for (III) : In this case by actual division reduce the fraction to the form f(x) +

(

)

c bx ax dx q px 2_{+ +} +

and then integrate.

Integrals of the form

Type III : (i)

ò

_a+bdx_sin2_x (ii)

ò

_{a bcos x}

dx

2

+ (iii)

ò

2 2

dx asin x bcos x+

(iv)

ò

₍

_{asinx bcosx}

₎

2

dx + (v)

ò

(

)

d x cos c x cos x sin b x sin a dx x tan 2 2 _{+ + +} f

where f (tan x) is a polynomial in tan x.

Rule : We shall always in such cases divide above and below by cos2x ; then put tan x = t i.e. sec2 x dx = dt then the question shall reduce to the forms

ò

₍

2

₎

dt at +bt c+ or

ò

( )

(

2

)

t dt at bt c j + + . form the of Integrals : IV Type _(i)

ò

x sin b a dx + (ii)

ò

a bcosx dx +

(iii)

ò

_asinx+dx_bcosx+c (iv)

ò

(

₍

p_acos_cosx_x+₊q_bsin_sinx_x

₎

)

dx (v)

ò

_ap_coscos_xx₊+_bq_sinsin_xx₊+r_c

Rule for (i), (ii), and (iii):

write cos x = 2 / x tan 1 2 / x tan 1 2 2 +

-, sinx = ₁₊2tan_tanx2_x/2_/2 The numberator shall become sec

2

x/2 and the denominator will be a quadratic in tan x/2. Putting tan x/2 = t i.e. sec2 x/2 dx = 2dt the question shall reduce to the form

ò

_at2₊dt_bt+c. Rule for (iv) : express the numberator as

I(Dr) + m(d.c. of Dr)

find l and m by comparing the coefficients of sin x and cos x and split the integral into sum of two integrals

as l

ò

dx + m

ò

r r D D of . c . d _{= lx + m ln | D}r _{| + c}

Rule for (v) : Express the numberator as

l(Dr) + m (d.c. of Dr) + n, find l, m, and n by comparing the coefficients of sin x, cos x and constant term and split the integral into sum of three integrals as

l

ò

dx + m

ò

r r D D of . c . d _{dx + n}

ò

_Dr dx _{= lx + m ln | D}r _{| + n}

ò

_Dr dx _{and to evaluate}

ò

_Dr dx _{proceed by the}

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Integrals of the form

Type V : (i)

ò

₍

(

4 2

)

4

₎

2 2 a kx x dx a x + + + (ii)

ò

₍

(

4 2

)

4

₎

2 2 a kx x dx a x + +

-where k is a constant, + ve, –ve or zero.

Rule for (i) and (ii) : Divide above and below by x2 then putting (i) t = x –

x a2 i.e., dt = çç_èæ + 2÷÷_øö 2 x a 1 _{dx and dt = ÷÷} ø ö çç è æ - ₂2 x a 1 _dx

then the questions shall reduce to the form

ò

2 2 c t dt + or

ò

2 2 c t dt -Remember : (i)

ò

4 2 4 2 a kx x dx x + + = 2 1

ò

₍

(

4 2

)

4

₎

2 2 x kx x dx a x + + + + 2 1

ò

₍

(

4 2

)

4

₎

2 2 a kx x dx a x + + -(ii)

ò

₍

₄ dx_{2 4}

₎

x +kx +a = _2a2 1

ò

₍

(

4 2

)

4

₎

2 2 x kx x dx a x + + + – _2a2 1

ò

₍

(

4 2

)

4

₎

2 2 a kx x dx a x + + -(iii)

ò

₍

₂

₎

n k x dx + = _k

(

_{2n 2}

)

(

_x2 _k

)

n1 x -+ - +

(

)

(

2n 2

)

k 3 n 2

-ò

₍

₂

₎

n1 k x dx -+ Integrals of the form

Type VI : (i)

ò

₍

_{) (}

₎

b ax B Ax dx + + (ii)

(

2

)

(

)

dx Ax +Bx C+ ax b+ (iii)

ò

_{( )}

(

2

)

dx Ax+B ax +bx+c (iv)

ò

₍

2

_{) (}

2

₎

dx Ax +Bx+c ax +bx c+

Rule for (i) and (ii) : Put ax + b = t2 Rule for (iii) : Put Ax + B =

t 1 Rule for (iv) : Put ax2₂ bx c

Ax Bx C

+ +

+ + = t

2

(iv) Integration by partial fraction :

Form of the Rational Function Form of the Partial Fraction

1.

₍

_x-px_a

₎₍

+_xq_-b

₎

, a ¹ b b x B a x A -+ -2.

₍

₎

2 a x q px -+ ;

₍

₎

2 a x B a x A -+ -3.

₍

_x-px_a

₎₍

2_x+_-qx_b

₎₍

+_xr_-c

₎

; c x C b x B a x A -+ -+ -4.

₍

_xpx _a

_{) (}

2qx_{x b}r

₎

2 -+ +

(

)

x b C a x B a x A 2 + -+ -5.

₍

₎

₍

₎

c bx x a x r qx px 2 2 + + -+ + ; c bx x C Bx a x A 2_{+ +} + +

-where x2 + bx + c cannot be factorised further.

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5.4 PROPERTIES OF DEFINITE INTEGRALS

(i)

ò

_aaf x dx 0( ) = _(ii) b ( ) af x dx

ò

= –

ò

_baf x dx( ) _. (iii)

ò

_abf x dx( ) ₌ c ( ) af x dx

ò

+

ò

_cbf x dx( ) _{(a < c < b) (iv)} b ( ) af x dx

ò

=

ò

_abf a b x dx( + - ) (v)

ò

₀af x dx( ) ₌ a ( ) 0f a x dx

-ò

(vi)

ò

₀2af x dx( ) ₌ a ( ) 0f x dx

ò

+ ( ) a 0f 2a x dx

-ò

(vii)

ò

₀2af x dx( ) _{= 2} a ( ) 0f x dx

ò

,if f(2a-x) = f(x) and

ò

₀2af x dx( ) _{= 0, if f(2a-x) = – f(x)}

(viii) a_af x dx( )

-ò

= 2

ò

₀af x dx( ) _{, if f(x) is even function, i.e., f(–x) = –f(x).}

5.5 INTEGRATION AS A LIMIT OF SUMS.

( ) b af x dx

ò

=

0

hLt® h[f(a) + f(a + h) + f(a + 2h) + ...+ f(a + n-1h)] or = (b – a)

¥ ® hLt n

1

[f(a) + f(a + h) + f(a + 2h) + ...+ f(a + n-1h)], where h =

n a b

-The following results are used for evalating questions based on limit of sums.

(i) 1 + 2 + 3 + ...+ (n – 1) =

å

(n–1) =

(

n 1 n

)

2 -(ii) 12 + 22 + 32 + ...+ (n–1)2 =

å

(n–1)2 = 6 ) 1 n 2 ( n ) 1 n ( - -(iii) 13 + 23 + 33 + ...+ (n – 1)3 =

å

(n–1)3 = 2 2 n ) 1 n ( úû ù êë é -(iv) a + ar + ...+ arn–1 =

[ ]

1 r 1 r a n -- _{(r ¹ 1).}

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Ex.1 Find f(x), if

(i) f’(x) = (x – 1)3 and f(2) + f(–2) = 0 (ii) f’(x) = a (cos x + sin x ), f(0) = 9 and fæ öç ÷_{è ø}p₂ = 15

Sol. (i) Here, f ’(x) = (x – 1) 3 \ f(x) =

ò

(x–1)3dx = 4 1 (x – 1)4 + C \ f(2) = 4 1 (2 – 1)4 + C = = 4 1

+ C Given that f(2) + f(–2) = 0, we have

Þ 4 1 _{+ C +} 4 81_{+ C = 0 Þ} 4 82 _{+ 2C = 0 Þ 20.5 + 2C = 0 Þ C = – 10.25}

Hence, from (1) we have

f(x) = 4 1 (x – 1)4 – 10.25 = 4 1 [(x – 1)4 – 41] (ii) Here, f ’(x) = a (cos x + sin x) \ f(x) =

ò

a (cos x + sin x)dx

= a

ò

cos x dx + a

ò

sin x dx = a sin x – a cos x + C ...(1)

\ f(0) = 9 = a×0 – a×1+C ...(2) Þ C–a=9 and f ÷ ø ö ç è æ p 2 =15 = a×1 – a × 0 + C Þ a+C =15

From (2) and (3), we have a = 3, C = 12 Hence, from (1), we have f(x)=3(sin x–cos x)+12

Ex.2 Integrate the following functions w.r.t. x : (i) (1 log x)2 x + _(ii)

(

x 1 x log x

)(

)

2 x + + (iii) x x 2+

Sol. (i) Put 1 + log x = y. Then,

x 1 dx = dy So, I =

ò

(

)

x x log 1+ 2 _dx Þ

ò

y2 dy = 3 y3 + C =

(

)

3 x log 1+ 3 _{+ C}

(ii) Put x + log x = y. Then, ÷ ø ö ç è æ + x 1 1 _{dx = dy Þ} ÷ ø ö ç è æ + x 1 x dx = dy So, I =

ò

(

)(

)

x x log x 1 x_{+ +} 2 dx

SOLVED PROBLESM

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=

ò

y2 dy = 3 y3 + C =

(

)

3 x log x₊ 3 + C

(iii) Put x + 2 = y. Then, dx = dy So, I =

ò

x _x+₂dy =

ò

(

)

2 1 y 2 y- dy Þ

ò

_{(y 2y}2₎ 1 2 3 - dy =

ò

2 1 2 3 y ) 2 dy y ( - dy Þ y C 3 2 . 2 y 5 2 ₂5_{- 3}2_{+ =}

₍

₎

₍

_{x 2}

₎

_C 3 4 2 x 5 2 2 3 2 5 + + -+

Ex.3 Find : (i)

ò

cos3x dx (ii)

ò

1 sin x dx -Sol. (i) We have cos3x =

4 1

(cos 3x + 3 cos x) \

ò

cos3 x dx =

4 1

ò

cos 3x dx + 4 3

ò

cos xdx = 4 1 3 x 3 sin ₊ 4 3_{sinx + C =} 12 1 _{sin3x +} 4 3_{sinx + C} (ii) We have x sin

1- = æ_èçcos2₂x+sin2₂x÷ö_ø-2sin₂xcosx₂

= 2 2 x sin 2 x cos ÷ ø ö ç è æ = 2 x sin 2 x cos

-\

ò

1-sinxdx =

ò

cos ₂xdx –

ò

sin ₂xdx = 2 sin ₂x + 2 cos ₂x + C

Ex.4 Find : (i)

ò

2 6 6 x x +a dx (ii)

ò

2 2 sec x tan x 4+ dx (iii)

ò

2 x 1 x 1 -- dx Sol. (i) Put x3= t so that 3x2 dx=dt or x2 dx =

3 1_dt \

ò

₆2 ₆ a x x - dx = 3 1

ò

₂

_{( )}

₃2 a t dt + =3 1_log 2 6 a t t+ + _{+ C =} 3 1_log 3 6 6 a x x + + _{+ C}

(ii) Put tan x = t so that sec2x dx = dt

\

ò

4 x tan x sec 2 2 + dx =

ò

2 d t t + 4 = log t₊ t2₊4 +C

= log tanx₊ tan2x₊4

+C (iii)

ò

1 x 1 x 2 -dx= 2 1

ò

x 1 x 2 2_- dx –

ò

1 x dx 2_- dx = x2_-1_-logx₊ x2₊1 + C

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Ex.5 Find : (i)

ò

2 x 2 x 2 x 3 + + + dx (ii)

ò

2 5 x 3 x 4 x 10 + + + dx

Sol. (i) Let x + 2 º A.

dx d

(x2 + 2x + 3) + B Þ x + 2 º A.(2x + 2 ) + B On comparing coefficients of like powers of x, we have

1 = 2A and 2 = 2A + B Þ A = 2 1 and B = 1 So,

ò

3 x 2 x 2 x 2_{+ +} + dx =

ò

(

)

3 x 2 x 1 2 x 2 2 1 2_{+ +} + + dx = ₂1

ò

3 x 2 x 2 x 2 2_{+ +} + dx +

ò

3 x 2 x dx 2_{+ +} = 2 1 ×2 _x2_{+2x+3+C1 +}

ò

(

)

₂

( )

2 2 1 x dx + + = x2₊2x₊3 + C 1+log x 1 x 2x 3 2_{+ +} + + _+C 2 = _x2_+2x+3 + log x+1+ x2+2x+3 + C (ii) Let 5x + 3 º A. dx d _(x2 _{+ 4x + 10) + B Þ 5x + 3 º A (2x + 4) + B}

On comparing coefficients of x, we have 5 = 2A and 3 = 4A + B Þ A =

2 5 and B = – 7 \

ò

+ + + 10 x 4 x 3 x 5 2 dx dx 10 x 4 x 7 ) 4 x 2 ( 2 5 2

ò

₊+ ₊-

ò

+ + -

ò

+ + + = 10 x 4 x dx 7 dx 10 x 4 x 4 x 2 2 5 2 2

( )

ò

+ + -+ + + ´ = 2 2 1 2 6 ) 2 x ( dx 7 C 10 x 4 x 2 2 5 =5 x2+4x+10+C1-7log(x+2)+ x2+4x+10+C2 C 10 x 4 x 2 x log 7 10 x 4 x 5 2+ + - + + 2+ + + = Ex.6 Find :

ò

_{( x 1)( x 2 )+} dx ₊ Sol. We write _(x+1)(1_x+2) º 1 x A + – x 2 B +

where real numbers A and B are to be determined suitabley. This gives 1 º A (x + 2) + B(x + 1)

Equation the coefficient of x and the constant terms, we get A + B = 0 and 2A + B = 1 Þ A = 1 and B = –1 Thus, _(x+1)(1_x+2) = 1 x 1 + + x 2 1 + - _Þ

ò

(x+1)(x+2) dx =

ò

ò

+ -+ x 2dx 1 dx 1 x 1

= log | x+1 | – log | x+2 | + C = log

2 x 1 x + -+C

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Ex.7 Find : (i) 2

3x 1 dx ( x 2 ) -+

ò

(ii) 2 x dx (x - 1) )(x + 2)

ò

Sol. (i) Let ₂

) 2 x ( 1 x 3 + - _º 2 x A + + (x 2)2 B + Then, 3x – 1 º A(x + 2) + B

Comparing coefficients of x and the constant terms on both sides, we get 3 = A and –1 = 2A + B Þ A = 3 and B = – 7 Thus,

ò

ò

ò

+ -+ = + -dx ) 2 x ( 7 dx 2 x 3 dx ) 2 x ( 1 x 3 2 2 = 3 log |x + 2| – 7 . (x 2) 1 + + C (ii) Let ) 2 x ( ) 1 x ( x 2 ₊ - º x 1 A - + (x 1)2 B - + x 2 C + Then x = A (x – 1) (x + 2) + B(x + 2) + C (x + 1)2 x = A (x2 _{+ x – 2) + B (x + 2) + C (x}2 _{– 2x + 1)}

Comparing coefficients of x2_{, x and the constant terms on both sides, we get} 0 = A + C; 1 = A + B – 2C ; 0 = –2A + 2B + C Þ A = 9 2 ; B = 3 1 ; C = – 9 2 Thus, dx ) 2 x ( ) 1 x ( x 2

ò

_{- +} =

ò

ò

ò

+ -+ - x 2dx 9 2 dx ) 1 x ( 3 1 dx 1 x 9 2 2 = 9 2 log |x – 1| – ) 1 x ( 3 1 - – 9 2 log |x + 2| + C = 9 2 log _xx₊-₂1 – _3(x1_-1) + C

Ex.8 Find : (i) _{x sin x dx}-1

ò

(ii) _{x cos x dx}-1

ò

Sol. (i) Put x = sin t so that d x = cos t . dt So,

ò

xsin-1xdx=

ò

sint.t.costdt

=

ò

t.sin2tdt 2 1 =

[

t.

ò

sin2tdt-

ò ò

[

1. sin2tdt

]

dt

]

2 1 = _êëé + _úûù -

ò

2 dt t 2 cos 2 t 2 cos . t 2 1 =– C 2 t 2 sin . 4 1 t 2 cos . t 4 1 _{+ +} = sint 1 sin t C 4 1 ) t sin 2 1 ( . t 4 1 _- 2 _{+ -} 2 ₊ = x 1 x C 4 1 x sin ) 1 x 2 ( 4 1 2_- -1 _{+ -} 2 ₊

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= -

ò

tsin2tdt 2 1 ₌ C 2 t 2 sin . 4 1 t 2 cos . t 4

1 _{- + (Using (1) give below)}

= 4 1 t(2 cos2 _{t – 1) –} 4 1 t cos 1- 2 . cos t + C = 4 1 (2x2 _{– 1) cos}–1 _{x –} 4 x ₂ x 1- + C

Ex.9 Find : (i) 1 - 1 ₂ dt

logx (logx) é ù ê ú ê ú ë û

ò

(ii) _ex sin4x - 4 _dx 1 - cos4x æ ö ç ÷ è ø

ò

Sol. (i) Let log x = t. Then x = et _and

x dx = dt \ dx ) x (log 1 x log 1 2

ò

_úú û ù ê ê ë é - = e dx t 1 t 1 t 2

ò

÷÷ ø ö çç è æ -=

ò

+ = t 1 ) t ( f where , dt ) t ( 'f ) t ( f [ et = et _{. f(t)+C} = et _. t 1 + C = _logx_x + C (ii)

ò

÷ ø ö ç è æ -x 4 cos 1 4 x 4 sin ex dx = _ex sin4x 4 1 cos4x 1 cos 4x é _- ù ê _{- -} ú ë û

ò

dx = dx x 2 sin 2 4 x 2 sin 2 x 2 cos x 2 sin 2 ex _{2 2}

ò

ú û ù ê ë é -=

ò

ex

[

cot2x-2cosec22x

]

dx =

ò

_{e [f(x) f '(x) dx}x + _{, where f(x) = cot 2x} = ex _{. f(x) + C = e}x _{. cot 2x + C}

Ex.10 Find : (i)

ò

x - 8x +7 dx2

(ii)

ò

1 - 4x - x dx2 Sol. (i) I =

ò

x2-8x+7dx =

ò

[

(x-4)2-9

]

dx =

ò

(x-4)2-32 dx = 2 4 x -7 x 8 x2- + – 2 9 _log 7 x 8 x ) 4 x ( - + 2- + + C (ii) I=

ò

1-4x-x2 dx=

ò

-(x2+4x-1)dx =

ò

-é_ë(x 2)+ 2-5 dxù_û =

ò

5-(x+2)2 dx = 2 1_{(x + 2)} 2 x x 4 1- - + ₂5sin–1 ÷÷_ø ö çç è æ + 5 2 x + C

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Ex.11 Evaluate te following definite integrals as limit of sums : 2

0x dx

ò

Sol. We have

ò

b

af(x) = h 0

lim

® h{f(a) + f(a + h) + f(a + 2h) +...+ f[a + (n – 1)h]}

where h =

n a b

-(i) Here, a = 0, b = 2, f(x) = x and h = n 0 2 = n 2 _{or nh = 2} \

_ò

2 0xdx = h 0 lim ® h{f(0) + f(0 + h) + f(0 + 2h) + ...+ f[0 + (n – 1)h]} =_hlim_®0 h{0 + (0+h) + (0+2h) +....+ [0 + (n–1)h]} = _hlim_®0 h[h + 2h +...+ (n – 1)h] =_hlim_®0 h2_{[1+2+....+ (n – 1)] =} hlim®¥ 2 n 2_÷ ø ö ç è æ úû ù êë é -2 n ) 1 n ( (Q h = n 2 _{and as h ® 0, n ® ¥)} = _nlim_®¥ 2 n 1 1 2 _ú= û ù ê ë é ÷ ø ö ç è æ

-Ex.12 Show that π /2

0 f(sin2x) sin x dx

ò

= ₂ π / 4 0 f(cos2x)cosx dx.

ò

Sol. Let x = 4 p – t. Then t = 4 p _{– x Þ dt = –dx} When x = 0, t = 4 p – 0 = 4 p and when x = 2 p , t = 4 p – 2 p = – 4 p _\

ò

0p 2/ f(sin2x)sinxdx =

ò

_p-p ç_èæ -p ÷_øö -þ ý ü î í ì ÷ ø ö ç è æ -p 4 / 4 / f sin2 4 t sin 4 t ( dt)

= /4_/4f(cos2t) sin cos t cos sin t dt

4 4 p -p p p æ _- ö ç ÷ è ø

ò

= /4_/4f(cos 2t) 1 cos t 1 sin t dt

2 2 p -p æ _- ö ç ÷ è ø

ò

= 1 /4_/4f(cos 2t)cos t dt 1 /4_/4f(cos 2t)sint dt

2 2 p p -p - -p

ò

ò

/4 0 1 1 .2 f (cos2t)cos t.dt . 0 2 2 p =

ò

-[Q f (cos 2t) cos t is an even function ; and f (cos 2t) sin t is an odd function]

/4 0

2 p f (cos 2t)cos t dt = -

ò

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Q.1. Evalute : (i)

ò

_- dx x cot x ec cos x ec cos (ii)

ò

+ dx x cos . x sin x cos x sin 2 2 6 6 (iii)

ò

1+cos3xdx Q.2 Evalute : (i) dx x 2 cos 1 x 2 sin tan1

ò

÷ ø ö ç è æ +

-(ii) tan 1(secx tanx)dx

ò

- + _,

2 x 2 –p< < p Q.3 (i) If f¢(x) = sin2x + cos3x + 5, f(0) =

2 3

, find (x)

(ii) If f¢(x) = a sinx + b cos x and f¢(0) = 4, f(0) = 3, ÷ ø ö ç è æ p 2 f _{= 5, find f(x)}

(iii) The gradient of a curve is given by 2x 3₂ x

= - . The curve passes through (1,1). Find its equation. (iv) The gradient of a curve is 6x2 _{– 2ax + a}2_{. The curve passes through the point (0, 0) and (1, 8). Find a.}

Q.4 Evaluate : (i)

ò

sinxsin2xsin3xdx _(ii)

ò

tanxtan2xtan3xdx Q.5 Evaluate : (i) sin(x a)dx

sin x

-ò

(ii) dx ) a x sin( x sin

ò

- (iii) sin(x a)sin(x b)dx 1

ò

- -Q.6 Evalute : (i) dx e e 1 x x

ò

₊ - (ii) dx x x 1 3 / 1 2 / 1

ò

+ Q.7 Evaluate : (i) dx ) x cos b a ( x 2 sin 2

ò

+ (ii) sin xsin(x a)dx

1

3

ò

₊ , a ¹ np, nÎz Q.8 Evalute : (i)

ò

_log(1+x)(1+x)_{dx (ii)}

ò

- dx x x sin 2 1

Q.9 Evaluate : (i)

ò

sec3xdx _{(ii) dx}

x cos x sin x cos x sin 1 1 1 1

ò

- -+

-Q.10 Evaluate : (i)

ò

sin(logx)dx (ii) eaxcosbxdx

ò

Q.11 Evaluate :

ò

+ -dx x 1 x 1 Q.12 Evaluate : (i)

ò

₊-_aa dx ) x ( sin ) x ( sin (ii)

ò

₊- dx x a x a

Q.13 Evaluate : (i)

ò

(3x-2) x2+x+1dx _(ii)

ò

(2x-5) x2-4x+3dx Q.14 Evaluate : (i) 1 dx 3 4cos x+

ò

(ii)

ò

_{+ +} dx x sin x cos 1 x cos (iii)

ò

_{+ +} dx x cos x sin 2 3 1 Q.15 Evaluate : (i)

ò

₊ dx x cot 1 1 (ii)

ò

+₊ dx x sin 2 x cos 3 x cos 2 x sin 3 Q.16 Evaluate : (i) dx ) 4 x ( ) 1 x ( 1 2 2

ò

+ + (ii) (x 1)(x 3)dx x 2 2 2

ò

+ + (iii) (x 1)(x 4)dx 2 x 2 2 2

ò

+ + + Q.17 Evaluate : (i)

ò

₊ dx 1 x 1 4 (ii)

ò

_{+ +} + dx 1 x x 1 x 2 4 2

(iii)

ò

( tanx+ cotx)dx

UNSOLVED PROBLEMS

EXERCISE – I

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Q.18 Evaluate : (i) dx ) x (log 1 x log 1 2 e e 2

ò

_þýü î í ì - _(ii) 1 0 1 x dx 1 x -+

ò

(iii)

ò

₀¥_(x2_+a2_)(x2_+b2₎ dx (iv)

ò

p + 2 / 0 2cosx 4sinxdx 1

Q.19 Evaluate as limit of sums : (i)

ò

_absinxdx _(ii) 4(x e )dx

0 x 2

ò

+ _(iii) 3(2x 3)dx 1

ò

+ _(iv) 3(3x 2x 1)dx 1 2

ò

+ +

Q.20 Prove that /2log(sinx)dx

0

ò

p = (log2) 2 dx ) x (cos log 2 / 0 p -=

ò

p Q.21 Evaluate : (i)

ò

p p -4 / 4 / 2_xdx sin _(ii)

ò

p p - + -2 / 2 / dx x sin 2 x sin 2 log Q.22 Evaluate :

ò

p p + 3 / 6 / dx x tan 1 1 Q.23 Evaluate : (i)

ò

p p -+ 2 / 2 / dx |) x | cos | x | (sin _(ii)

ò

-1 1 | x | _dx e _(iii) |x 2x 3|dx 2 0 2

ò

+ -(iv)

ò

4 1 dx ) x ( f , where f(x) = |x – 1| + |x – 2| + |x – 3| Q.24 Evaluate : (i)

ò

4 1 dx ) x ( f , where f(x) = î í ì < £ -£ £ + 4 x 2 when , 2 x 9 2 x 0 when , 4 x 3 2 (ii)

ò

-1 0 dx | 3 x 5 | _(iii)

ò

p 0 dx | x cos | _(iv)

ò

p 2 0 dx | x sin | Q.25 Evaluate : (i)

ò

p + 2 / 0 2 dx x cos x sin x sin (ii)

ò

p + 0 dx x cos x sin x (iii)

ò

p + 0 dx x tan x sec x tan x (iv)

ò

p + 2 / 0 dx x cos x sin x (v)

ò

p + 0 2_xdx cos 1 x sin x Q.26 Evaluate : (i)

ò

p + 2 / 0 3 3 3 dx x cos x sin x sin (ii)

ò

p + 2 / 0 dx x cos x sin x sin (iii)

ò

p + 4 / 0 dx ) x tan 1 ( log (iv)

ò

p + -2 / 0 dx x cos x sin 1 x cos x sin (v)

ò

-1 0 n_dx ) x 1 ( x _(vi)

ò

p 2/ 0 dx ) x (tan log x 2 sin

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Q.1 (i) Evaluate dx x cos x sin x sin 2 / 0 7 7 7

ò

p + . (ii) (x 2x)dx 3 0 2

ò

+ as limit of a sum. [C.B.S.E. 2000]

Q.2 (i) dx 3 x 4 x 1 x 2 2

ò

₊+ ₊ (ii)

ò

sin4xdx. (iii) dx x cos x x 2 sin 1 2

ò

₊+ (iv)

ò

xsin-1xdx. [C.B.S.E. 2000]

Q.3 (i)

ò

sin7xsinxdx. (ii) dx 3 x 2 x 2 3 x 4 2

ò

₊+ _- (iii)

ò

log(1+x2)dx. (iv) dx ) 2 x )( 1 x ( 1 x

ò

+ -- _. [C.B.S.E. 2001] Q.4 (i) Evaluate as the limit of a sum (x 3)dx

2 0 2

ò

+ . (ii) Evaluate dx x sin 1 x 0

ò

p + . (iii) 1 cos xdx x sin x 0 2

ò

p + . [C.B.S.E. 2001]

Q.5 (i)

ò

eaxcosbxdx. (ii) dx ) 1 x )( 1 x ( x 2

ò

_{+ +} (iii)

ò

_{+ + +} 2 x 1 x dx (iv)

ò

+ -4x 8 x dx 2 . [C.B.S.E. 2002]

Q.6 (i) Evaluate cos xdx

2 / 0 2

ò

p

. (ii) Prove that

ò

+ = p

p 2 dx ) x cot x tan ( 2 / 0 .

(iii) Prove log2.

8 d ) tan 1 ( log 4 / 0

ò

p p = q q + (iv) Evaluate dx x 1 x 1 e 2 1 2 x

ò

ê_ëé - ú_ûù . [C.B.S.E. 2002]

Q.7 (i)

ò

sec3x (ii) dx

) x 1 ( x tan 2 1

ò

₊- . (iii)

ò

- ₊ -x sin 1 x sin 1 tan 1 dx. (iv)

ò

_(x+2x₎₍dx_3-2x). [C.B.S.E. 2003]

Q.8 (i) sin2x.log(tanx)dx 0.

2 / 0 =

ò

p

(ii) Prove that dx x a x a a a

ò

- + - _{= ap.}

(iii) Evaluate cos2x.logsinxdx

2 / 4 /

ò

p p . (iv) dx x cos x sin x sin 2 / 0 2

ò

p + [C.B.S.E. 2003]

BOARD PROBLES

EXERCISE – II

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Q.9 (i)

ò

(sin-1x)2dx. (ii)

ò

sin42xdx. (iii)

ò

+ +2sinx cosx 3 dx _(iv) dx e e 4 5 e x 2 x x

ò

_{- -} . [C.B.S.E. 2004] Q.10 (i) Evaluate f(x)dx 0 5

ò

-, where f(x) = |x| + |x + 3| + |x + 6|. (ii) (x x)dx 2 0 2

ò

+ as a limit of sums (iii)

ò

p

+

05 4cosx

dx

[C.B.S.E. 2004]

Q.11 (i) Prove that 1.

2 dx x 1 x 1 1 0 -p = +

-ò

(ii) 1 sin2x dx 2 1. 4 / 0 -=

-ò

p

(iii) Evaluate 2tan xdx

4 / 0 3

ò

p . (iv)

ò

p + + 2 / 0 (1 sinx)(2 sinx) dx x cos . [C.B.S.E. 2004] Q.12 (i) dx x cos b x sin a x 2 sin 2 2 2 2

ò

₊ . (ii) dx x ) x (log 16 2

ò

+ . (iii) dx ) 3 x )( 2 x )( 1 x ( 1 x 2

ò

- + -- _. (iv) dx 12 x 6 x x 2 2

ò

_{+ +} (v) dx ) x cos b a ( x 2 sin 2

ò

₊ (vi) dx ) x (log 1 x log 1 2

ò

_úú û ù ê ê ë é - . [C.B.S.E. 2005] Q.13 (i)

ò

+₊ 2 x 2 ) 1 x ( e ) 1 x ( dx. (ii)

ò

(x+3) 3-4x-x2 dx. (iii) dx 3 x 4 x 2 1 x 2 2

ò

₊+

-(iv)

ò

tanqdq (v)

ò

cos4xdx (vi) dx

) 1 x )( 2 x ( 1 x x 2 2

ò

₊ + +₊ . [C.B.S.E. 2006] Q.14 (i) Evaluate

ò

p p + 3 / 6 / dx x tan 1

1 _{. (ii) Prove that}

. dx ) x a ( f dx ) x ( f b a a 0

ò

ò

= - Hence evaluate

ò

p + 2 / 0 1 tanx dx . [C.B.S.E. 2006]

Q.15 (i)

ò

sin4xsin8xdx (ii) dx

) x 2 cos 2 )( x 2 cos 1 ( x 2 sin

ò

- - (iii) 1 x dx x 1 4 2

ò

₊ -[C.B.S.E. 2007] Q.16 (i) (2x 3x 5)dx 3 0 2

ò

+ + as limit of sums (ii) dx

x ec cos x sec x tan x 0

ò

p (iii) x 2 x dx 2 0

ò

- as limit of sums [C.B.S.E. 2007] Q.17 (i) dx 5 x 6 x x 4 x 2 3 2

ò

₊ + ₊ (ii) If

ò

(eax+bx)dx = 4 e4x ₊ 2 x

3 2 _{find the value of a and b}

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Q.18 (i) dx x tan x sec x tan x 0

ò

p

+ (ii)

ò

₁₊x_cossin2x_xdx (iii) _{sinx cosx} dx

x 2 / 0

ò

p + (iv) cot (1 x x )dx 0 2 1

ò

p - _{- +} [C.B.S.E. 2008] Q.19 (i) dx x tan 3 x sec2

ò

+ (ii) x x 2x e dx 5 4e- -e

ò

(iii) dx ) 2 x ( e ) 4 x ( 3 x

ò

-

-(iv)

ò

xsin-1xdx [C.B.S.E. 2009]

Q.20 (i) dx e e e 0 cosx cosx x cos

ò

p

-+ (ii) (2logsinx logsin2x)dx

2 / 0

ò

p - (iii) dx x sin b x cos a dx x 0 2 2 2 2

ò

p + [C.B.S.E. 2009]

Q.21 (i)

ò

sec2(7-4x)dx (ii) x 2 dx

(x 2) (x 3) + -

-ò

[C.B.S.E. 2010] Q.22 (i) sin xdx 2 / 2 / 5

ò

p p -(ii) dx 3 x 4 x x 5 2 1 2 2

ò

_{+ +} [C.B.S.E. 2010] Q.23 6x 7 dx (x 5)(x 4) + -

-ò

[C.B.S.E. 2011] Q.24 Evaluate : (i) /2 0 x sin x dx 1 cos x p + +

ò

(ii) /3 /6 dx 1 tan x p p

ò

+ [C.B.S.E. 2011] Q.25 Evaluate (i)

ò

-2 1 3 _{x dx} x _{. (ii)}

ò

p + 0 2_xdx cos 1 x sin x [C.B.S.E. 2012]

Q.26 Evaluate (i) x₁sin_x2x dx 1

ò

_-- (ii) _(x x₁₎2_(x1 ₃₎ dx 2 + -+ [C.B.S.E. 2012] Q.27 Evaluate : [C.B.S.E. 2013]

ò

- a a -cos x cos 2 cos x 2 cos dx Evaluate :

ò

_x2x₊+₂2_x+3 dx Q.28 Evaluate : [C.B.S.E. 2013]

(

)

ò

_xxdx5 _{+ 3} Q.29 Evaluate : [C.B.S.E. 2013] dx e 1 1 2 0 sinx

ò

p ₊

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ANSWER KEY

EXERCISE – 1 (UNSOLVED PROBLEMS) 1. (i) –cot x – cosec x + c (ii) tan x – cot x – 3x+c (iii)

3 2 2 _sin 2 x 3 _{+ c 2. (i)} 2 x2 _{+c (ii)} 4 x p + 4 x2 _{+ c} 3. (i) – 2 x 2 cos + 3 x 2 sin

+5x+2 (ii) 2 cos x + 4 sin x + 1 (iii) y = x2 +

x 3 – 3 (iv) 3, –2 4. (i) – 16 x 4 cos – 8 x 2 cos + 24 x 6 cos + c (ii) – 3 1 log cos 3x + 2 1 log cos 2x + 2 1 log cosx + c

5. (i) x cos a – sin a log sin x+c (ii) sin a log sin (x – a)+(x – a) cos a (iii) cosec (a – b) log _sin(sin(x_x--_ba₎) + c

6. (i) tan–1 x e + c (ii) 2 _x – 3x1/3 + 6x1/6 – 6log (x1/6 + 1) + c 7. (i) ₂

b 2 -úû ù êë é + + x cos b a a ) x cos b a log( _{+ c}

(ii) –2 cosec a cosa+cotx+sina 8. (i) 2 2 x ) 1 x ( + log (x + 1) – 4 x2 _– 2 x + c (ii) – x x sin-1 _{+ log} 1 1 x 2 x é _{- -} ù ê ú ê ú ë û

9. (i) sec x tan x

2 ´ + 2 1 log tan ÷ ø ö ç è æ +p 2 x 2 + c (ii) p 2 (2x – 1) sin–1 _x + _p2 _x-x2 – x + c 10. (i) 2

x_{[sin(log x) – cos (log x)] + c (ii)}

2 2 ax b a e + [a cos bx + b sin bx] + c 11. (i) –2 _1-x + _{x 1-x} – sin–1 _x + c

12. (i) – cos a sin–1 ÷ ø ö ç è æ a cos x cos

– sin a log [sin x + _sin2_x-sin2_{a] + c (ii) a sin}–1

a x + _a2_-x2 _{+ c} 13. (i) (x2 + x + 1)3/2 – 16 21 1 x x 8 ) 1 x 2 ( 7 + 2_{+ + - log ú} û ù ê ë é + + + ÷ ø ö ç è æ + x x 1 2 1 x 2 _{+ c} (ii) 3 1_(x2 _{– 4x + 3)}3/2 _– 2 1_{(x – 2)} 3 x 4 x2- + – ₂1 log [(x – 2) + x2-4x+3] 14. (i) 1₇ log _úú ú û ù ê ê ê ë é + + -+ 7 4 2 x tan 3 7 4 2 x tan 3 + c (ii) c 2 x tan 1 x sin x cos 1 log 2 1 2 x ₊ ú ú ú û ù ê ê ê ë é + + + + (iii) tan–1 ÷ ø ö ç è æ _{+ 1} 2 x tan + c 15. (i) 2 x – 2 1

log (sin x + cos x) + c (ii)

13 x 12 – 13 5

log (3 cos x + 2 sin x) + c

16. (i) 3 1 tan–1x – 6 1 tan–1 ÷ ø ö ç è æ 2 x (ii) 2 1 log ÷÷_ø ö ç ç è æ + + 3 x 1 x 2 2 + c

17. (i) ₂1₂tan x _2x1 ₄1₂log x_x2 2_2xx 1₁ c 2 2 1 ₊ ú ú û ù ê ê ë é + + + -÷ ÷ ø ö ç ç è æ

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(ii) c x 3 1 x tan 3 1 1 2 ₊ ú ú û ù ê ê ë é

-- _(iii) 1 _tan 1 tan x 1 _c

2 2 tan x - æ - ö₊ ç ÷ è ø 18. (i) e 2 e2 _{- (ii) 1} 2 -p

(iii) _2ab(ap_+b) (iv) ₂1₂log

ú ú û ù ê ê ë é -+ ) 2 5 ( 2 1 5

19. (i) cos a–cos b (ii)

2 e

15+ 8 _{(iii) 14 (iv) 36 21. (i)} 4 p – 2 1 _{(ii) 0 22.} 12 p

23. (i) 4 (ii) 2e–2 (iii) 4 (iv) 2

19 _{24. (i) 66 (ii)} 10

13 _{(iii) 2 (iv) 4 25. (i) –} 2 1 log ( ₂ – 1) (ii) p (iii) p ÷ ø ö ç è æ -p 1 2 (iv) 4 2 p log _÷÷ ø ö ç ç è æ -+ 1 2 1 2 (v) 4 2 p _{26. (i)} 4 p (ii) 4 p (iii) log2 8 p (iv) 0 (v) ) 2 n )( 1 n ( 1 + + (vi) 0

EXERCISE – 2 (BOARD PROBLEMS)

1. (i) 4 p

(ii) 18 2. (i) 2 x2+4x+3-3log_êëé(x+2)+ x2+4x+3ù_úû+c (ii)

32

1 _{[12x – 8 sin 2x+sin}

4x]+c

(iii) log (x + cos2x)+c (iv)

2 x2 _sin–1 x + 4 x 2 x

1- – ₄1sin–1x + c 13. (i) p/2 (ii) p/12 16. (i) p 1 2 p æ _- ö ç ÷ è ø (ii) 4 2 p _(iii) 2 2 p

log ( 2 + 1) (iv) ₂p – log 2 17. (i) 93₂ (ii) ₄

2 p _(iii) 15 16 2 18. (i)₂p (ii)₄p (iii) ₅3₂ (iv) 3 20 3. (i) 16 x 8 sin 12 x 6 sin + c (ii) 2 _2x2 _{2x 3} 1 _{log x} 1 _x2 _x 3 2 2 2 é ù + - + ê + + + - ú ê ú ë û

(iii) x log (1 + x2) – 2(x – tan–1x) + c (iv)

3 2 log(x + 1) + 3 1 log (x – 2) + c 4. (i) 3 26 _{(ii) p (iii)} 4 2 p 5. (i) ₂ ax ₂ b a e

+ [a cos bx + b sin bx] + c (ii) 4

1 _{log (x}2 _{+ 1) +} 2 1_tan–1_{x –} 2 1_{log (x + 1)} (iii) 3 2_{log (x + 2)}3/2 _– 3 2_{log(x + 1)}3/2 _{+ c (iv)} 2 1_tan-1 x 2 2 -æ ö ç ÷ è ø 6. (i) 4 p (iv) e 2 e2

-7. (i) sec x tan x + log (secx + tan x) (ii) x 1 x tan 1 + -+ 2 1 [log (1 + x) + tan–1x – 2 1 log (1 + x2)] + c (iii) 4 x p – 4 x2_{+ c} (iv) – 7 2 log (x + 2) – 14 3 (3 – 2x) + c 8. (iii) 4 1 _÷ ø ö ç è æ _-p₊ 2 log 2 1 _(iv) 2 1 log ( 2+1)

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9. (i) x(sin–1 x)2 + 2 _1-x2 _sin–1_{x – 2x + c (ii)}

8 1

[3x – sin 4x +

8 1

sin 8x] + c (iii) tan–1 ÷ ø ö ç è æ + 2 x tan 1 + c (iv) sin–1 _÷÷ ø ö ç ç è æ ₊ 3 2 ex + c 10. (i) 2 73 (ii) 3 14 (iii) 3 p

11. 1. (iii) 1–log 2 (iv) log

3 4 12. (i) _{2 2} b a 1 - log [a 2

sin2x + b2 cos2 x] + c (ii)

2 ) x (log 16 x log + 2 _{+ log[log x + 2} ) x (log 16+ ] (iii) – 6 1 log(x – 1) – 3 1 log(x + 2) + 2 1

log (x – 3) (iv) x – 3log (x2 + 6x + 12) + 6₃ + tan–1 ÷÷ ø ö çç è æ + 3 3 x + c (v) – ₂ b 2 úû ù êë é + + + x cos b a a ) x cos b a

log( + c (vi) _logx_x + c 13. (i) ex ÷ ø ö ç è æ + -1 x 1 x + c (ii) – 3 1_{[3 – 4x – x}2_]3/2₊ 2 ) 2 x ( + 2 x x 4 3- - +7₂ sin–1çç_èæ +x ₇2_øö÷÷+ c (iii) 2 1_{log [2x}2 _{+ 4x – 3] –} 5 4 2 log ú ú û ù ê ê ë é + + -+ 5 ) 1 x ( 2 5 ) 1 x ( 2 (iv) 1₂tan–1 _úú û ù ê ê ë é q -q tan 2 1 tan2 +₂1_{2 ú}ú û ù ê ê ë é + q + q + q -q 1 tan 2 tan 1 tan 2 tan 2 2 + c (v) 8 x 3 ₊ 32 1 _{sin 4x +} 4

1 _{sin 2x + c (vi) –2 log (x + 1) –} 1 x 1 + + 3 log (x + 2) + c 15. (i) 24 1

(3 sin x – sin 12x) + c (ii)

2 1 log ÷ ø ö ç è æ -x 2 cos 2 x 2 cos 1 + c (iii)₂-1₂log ú ú û ù ê ê ë é + + + -1 x 2 x 1 x 2 x 2 2 + c 17. (i) 3 1_{log (x}3 _{+ 1) + c (ii) a = 4, b = 3} 23. 6. x 9x 20 c 2 9 x 2 log . 34 20 x 9 x2 ÷+ 2 - + + ø ö ç è æ -+ + -25. (i) 11₄ (ii) p₄2 26. (i) sin1x

(

1 x2

)

x c + + -- - _(ii) C ) 1 x ( 1 . 2 1 | 3 x | log 8 5 | 1 x | log 8 3 ₊ -+ +