 Mathematics Some Properties of Definite Integrals For CBSE-NCERT
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color{red}♦ Some Properties of Definite Integrals

### Some Properties of Definite Integrals

We list below some important properties of definite integrals. These will be useful in evaluating the definite integrals more easily.

color {red} {P_0 : int_a^b f(x) dx = int_a^b f(t) dt}

color {red} {P_1 : int_a^b f(x) dx = -int_b^a f(x) dx . (In particular , int_a^a f(x) dx = 0)

color {red} {P_2 : int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx}

color {red} {P_3 : int_a^b f(x) dx = int_a^b f(a +b -x ) dx}

color {red} {P_4 : int_0^a f(x) dx = int_0^a f(a -x )dx}

(Note that P_4} is a particular case of P_3)

color {red} {P_5 : int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a -x ) dx}

color {red} {P_6 : {tt((int_0^(2a) f(x) dx = 2 int_0^a f(x) dx, if f(2a − x) = f( x)),(0,if f(2a -x) = - f(x)))

color {red} {P_7 : {tt ((int_(-a)^a f(x) dx = 2 int_0^a f(x) dx, "if f is an even function"),( 0 , "if f is an odd function"))

"Proof:" of P_0 It follows directly by making the color{green}{"substitution x = t."}

"Proof" of color {red} {P_1} Let F be anti derivative of f. Then, by the second fundamental theorem of calculus, we have

int_a^b f(x) dx = F(b) - F(A) =- [F(A) - F(b) ] = - int_b^a f(x) dx

Here , we observe that , if a = b , then int_a^b f(x) dx = 0

color {red}{ "Proof" } of P_2 Let F be anti derivative of f. Then

int_a^b f(x) dx = F(b) - F(a) ..................(1)

int_a^c f(x) dx = F(c) - F(a) ..................(2)

and int_c^b f(x) dx = F(b) - F(c) .......(3)

Adding (2) and (3), we get int_a^c f(x) dx + int_c^b f(x) dx = F(b) - F(a) = int_a^b f(x) dx

This proves the property P_2

"Proof" of P_3 Let t = a + b – x,

Then dt = – dx. When x = a, t = b and when x = b, t = a.

int_a^b f(x) dx = - int_b^a f( a +b - t) dt

= int_b^a f( a + b - t ) dt

= int_a^b f( a + b - x ) dx  by P_0

"Proof" of P_4 Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in P_3

"Proof" of P_5 Using P_2, we have int_0^(2a) f(x) dx = int_0^a f(x ) dx + int_a^(2a) f(x) dx

Let  t =2a -x in the second integral on the right hand side. Then

dt = – dx. When x = a, t = a and when x = 2a, t = 0. Also x = 2a – t.

Therefore, the second integral becomes

int_a^(2a) f(x) dx = - int_a^0 f(2a -t) dt = int_0^a f( 2a -t ) dt = int_0^a f( 2a -x ) dx

Hence int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a -x ) dx

"Proof" of color {red} {P_6} Using color {red} {P_5}, we have

int_0^(2a) f(x) dx = int_0^(2a) f(x) dx + int_0^a f( 2a -x ) dx ......(1)

Now , if f (2a – x) = f (x), then (1) becomes

int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(x) dx = 2 int_0^a f(x) dx

and if f (2a – x) = – f (x), then (1) becomes

int+0^(2a) f(x) dx = int_0^a f(x) dx - int_0^a f( x) d = 0

"Proof" of color {red} {P_7} Using color {red} {P_2}, we have

int_(-a)^a f(x) dx = int_(-a)^0 f(x) dx + int_0^a f(x), Then

t =- x  in the first integral on the right hand side.

dt = – dx. When x = – a, t = a and when

x = 0, t = 0. Also x = – t.

Therefore int_(-a)^a f(x) dx = - int_a^0 f(-t) dt + int_0^a f(x) dx

= int_0^a f(-x) dx + int_0^af(x) dx  ( by P_0 ) .................(1)

(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes

int_(-a)^a f(x) dx = int_0^a f(x) + int_0^a f(x) dx = 2 int_0^a f(x) dx

(ii) If f is an odd function, then f (–x) = – f (x) and so (1) becomes

int_(-a) f(x) dx = - int_0^a f(x) dx + int_0^a f(x) dx = 0
Q 3125580461 Evaluate  ∫_(-1)^2 | x^3 -x | dx
Class 12 Chapter 7 Example 30 Solution:

We note that x^3 – x ≥ 0 on [– 1, 0] and x^3 – x ≤ 0 on [0, 1] and that
x^3 – x ≥ 0 on [1, 2]. So by P_2 we write

 ∫_(-1)^2 | x^3 -x | dx = ∫_(-1)^0 ( x^3-x) dx + ∫_(0)^1 - (x^3 -x) dx + ∫_(1)^2 (x^3 -x) dx

= ∫_(-1)^0 (x^3 -x ) dx + ∫_(0)^1 (x-x^3 ) dx + ∫_(1)^2 (x^3 -x ) dx

 = [ x^4/4 - x^2/2]_(-1)^0 + [ x^2/2 - x^4/4]_(0)^1 + [x^4/4 - x^2/2]_(1)^2

= - (1/4 -1/2) + (1/2 -1/4) + (4-2) - (1/4 - 1/2 )

 = -1/4 +1/2 +1/2 -1/4 +2 -1/4 +1/2 = 3/2 - 3/4 +2 = 11/4
Q 3185180967 Evaluate  ∫_(-pi/4)^(pi/4) sin^2 x dx

Class 12 Chapter 7 Example 31 Solution:

We observe that sin^2 x is an even function. Therefore, by P_7 (i), we get

 ∫_(-pi/4)^(pi/4) sin^2 x dx = 2 ∫_(0)^(pi/4) sin^2 x dx

 = 2 ∫_(0)^(pi/4) ( 1- cos 2x)/2 dx = ∫_(0)^(pi/4) (1- cos 2x ) dx

= [ x-1/2 sin 2x ]_(0)^(pi/4) = (pi/4 -1/2 sin ( pi/2)) -0 = pi/4 - 1/2
Q 3115191060 Evaluate  ∫_(0)^pi (x sin x )/( 1+ cos^2 x) dx
Class 12 Chapter 7 Example 32 Solution:

Let I = ∫_(0)^pi (x sin x)/(1+cos^2 x) dx .Then, by P_4, we have

I = ∫_(0)^pi ( (pi-x) sin (pi- x) dx )/(1+ cos^2 (pi-x) )

 =∫_(0)^pi ( (pi-x) sin x dx)/(1+cos^2 x) = pi ∫_(0)^pi (sin x dx)/(1+ cos^2 x) -I

or 2I= pi ∫_(0)^pi (sin x dx)/(1+ cos^2 x)

or I= pi/2 ∫_(0)^pi (sin x dx)/( 1+ cos^2 x)

Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = – 1.
Therefore, (by P_1 ) we get

I = (-pi)/2 ∫_(1)^(-1) (dt)/( 1+t^2) = pi/2 ∫_(-1)^1 (dt)/(1+t^2)

= pi ∫_(0)^1 (dt)/(1+t^2) (by P_7 , since 1/(1+t^2) is even function)

= pi [ tan^(-1) t ]_(0)^1 = pi [ tan^(-1) 1- tan^(-1) 0 ] = pi [ pi/4 -0 ] = pi^2/4
Q 3155191064 Evaluate  ∫_(-1)^1 sin^5 x cos^4 x dx

Class 12 Chapter 7 Example 33 Solution:

Let I = ∫_(-1)^1 sin^5 cos^4 x dx . Let f(x) = sin^5 x cos^4 x . Then

f (– x) = sin^5 (– x) cos^4 (– x) = – sin^5 x cos^4 x = – f (x), i.e., f is an odd function.
Therefore, by P_7 (ii), I = 0
Q 3185191067 Evaluate  ∫_(0)^(pi/2) (sin^4 x)/(sin^4 x + cos^4 x) dx
Class 12 Chapter 7 Example 34 Solution:

Let I = ∫_(0)^(pi/2) (sin^4 x)/(sin^4 x+ cos^4 x) dx ..........(1)

Then, by P_4

I = ∫_(0)^(pi/2) (sin^4 (pi/2-x) )/( sin^4 (pi/2-x) + cos^4 (pi/2 -x) ) dx= ∫_(0)^(pi/2) (cos^4 x)/( cos^4 x + sin^4 x) dx .......(2)

Adding (1) and (2), we get

2I = ∫_(0)^(pi/2) (sin^4 x + cos^4 x)/( sin^4 x +cos^4 x) dx = ∫_(0)^(pi/2) dx = [x]_(0)^(pi/2) = pi/2

Hence  I = pi/4
Q 3125291161 Evaluate  ∫_(pi/6)^(pi/3) (dx)/(1+ sqrt (tan x) )
Class 12 Chapter 7 Example 35 Solution:

Let I = ∫_(pi/6)^(pi/3) (dx)/(1+ sqrt (tan x) ) = ∫_(pi/6)^(pi/3) ( sqrt (cos x) dx )/(sqrt (cos x) + sqrt (sin x) ) .........(1)

Then, by P_3   I = ∫_(pi/6)^(pi/3) (sqrt (cos (pi/3+pi/6 -x) ) dx)/( sqrt (cos (pi/3 + pi/6 -x) ) + sqrt (sin (pi/3+pi/6 -x) ) )

= ∫_(pi/6)^(pi/3) (sqrt (sin x) )/( sqrt (sin x) + sqrt (cos x) ) dx ...........(2)

Adding (1) and (2), we get

2I= ∫_(pi/6)^(pi/3) dx = [x]_(pi/6)^(pi/3) = pi/3- pi/6 = pi/6. Hence I = pi/12
Q 3115291169 Evaluate  ∫_(0)^(pi/2) log sin x dx
Class 12 Chapter 7 Example 36 Solution:

Let I = ∫_(0)^(pi/2) log sin x dx

Then, by P_4

I = ∫_(0)^(pi/2) log sin (pi/2-x) dx = ∫_(0)^(pi/2) log cos x dx

Adding the two values of I, we get

2I = ∫_(0)^(pi/2) (log sin x + log cos x ) dx

= ∫_(0)^(pi/2) ( log sin x cos x + log 2 - log 2 ) dx (by adding and subtracting log2)

 = ∫_(0)^(pi/2) log sin 2x dx-∫_(0)^(pi/2) log 2 dx (Why?)

Put 2x = t in the first integral. Then 2 dx = dt, when x = 0, t = 0 and when x = pi/2

t = pi

Therefore 2I = 1/2 ∫_(0)^pi log sin t dt - pi/2 log 2

=2/2 ∫_(0)^(pi/2) log sin t dt -pi/2 log 2  [by P_6 as sin (π – t) = sin t)

= ∫_(0)^(pi/2) log sin x dx - pi/2 log 2 (by changing variable t to x)

 = I - pi/2 log 2

Hence  ∫_(0)^(pi/2) log sin x dx = (-pi)/2 log 2
Q 3145391263 Find  ∫ cos 6x sqrt (1+ sin 6x) dx
Class 12 Chapter 7 Example 37 Solution:

Put t = 1 + sin 6x, so that dt = 6 cos 6x dx

Therefore  ∫ cos 6x sqrt (1+ sin 6x) dx = 1/6 ∫ t^(1/2) dt

= 1/6 xx 2/3 (t)^3/2 +C = 1/9 (1+ sin 6x )^3/2 +C
Q 3186134077 Find  ∫ ( (x^4 -x )^(1/4))/(x^5 ) dx
Class 12 Chapter 7 Example 38 Solution:

We have  ∫ ( (x^4 -x)^(1/4) )/x^5 dx = ∫ ( (1-1/x^3)^(1/4) )/x^4 dx

put 1-1/x^3 = 1-x^(-3) = t  , so that 3/x^4 dx = dt

Therefore  ∫ ( (x^4 - x)^(1/4) )x^5 dx = 1/3 ∫ t^(1/4) dt = 1/3 xx 4/5 t^(5/4) +C = 4/15 (1- 1/x^3)^(5/4) +C
Q 3136812772 Find  ∫ ( (x^4 -x)^1/4)/x^5 dx
Class 12 Chapter 7 Example 38 Solution:

We have  ∫ ( (x^4 -x)^1/4)/x^5 dx = ∫ ( (1-1/x^3 )^1/4 )/x^4 dx

put 1-1/x^3 = 1 - x^(-3) = t , so that 3/x^4 dx = dt

Therefore  int ( (x^4 -x)^1/4)/x^5 dx = 1/3 ∫ t^1/4 dt= 1/3 xx 4/5 t^5/4 +C = 4/15 (1- 1/x^3)^5/4 +C
Q 3176234176 Find  ∫ (x^4 dx)/( (x-1)( x^2 +1) )
Class 12 Chapter 7 Example 39 Solution:

We have

x^4/( (x-1) (x^2 +1) ) = (x+1) + 1/( x^3 - x^2 +x -1)

= (x+1) + 1/( (x+1) (x^2 +1) ) ........(1)

Now express 1/( (x-1) ( x^2 +1) ) = A/(x-1) + (Bx+C)/(x^2 +1) ......(2)

1 = A (x^2 + 1) + (Bx + C) (x – 1)

= (A + B) x^2 + (C – B) x + A – C

Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,

which give A = 1/2 , B=C = -1/2 Substituting values of A, B and C in (2), we get

1/( (x-1) (x^2 +1) ) = 1/(2 (x-1) ) -1/2 x/(x^2 +1) - 1/(2 (x^2 +1) ) .......(3)

Again, substituting (3) in (1), we have

x^4/( (x-1) (x^2 +x+1) ) = (x+1) +1/(2 (x-1) ) -1/2 x/( x^2+1) -1/(2 (x^2 +1) )

Therefore

 ∫ x^4/( (x-1) ( x^2 +x+1) ) dx = x^2/2 + x +1/2 log | x-1| -1/4 log (x^2 +1) -1/2 tan^(-1) x+C
Q 3126123071 Find  ∫ (x^4 dx)/( (x-1) (x^2+1) )
Class 12 Chapter 7 Example 39 Solution:

We have

x^4/( (x-1) ( x^2 +1) ) = (x+1) + 1/(x^3 - x^2 + x-1)

= (x+1) + 1/( (x+1) (x^2 +1) ) ............(1)

Now express 1/( (x-1)(x^2+1) ) =A/(x-1) + (Bx+C)/(x^2 +1) .......(2)

1 = A (x^2 + 1) + (Bx + C) (x – 1)

= (A + B) x^2 + (C – B) x + A – C

Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,

which give A =1/2 , B=C =-1/2 Substituting values of A, B and C in (2), we get

1/( (x-1) (x^2 +1) ) = 1/(2 (x-1) ) -1/2 x/(x^2 +1) -1/(2 (x^2 +1) ) ..........(3)

Again, substituting (3) in (1), we have

x^4/( (x-1) (x^2 +x+1) ) = (x+1) +1/(2 (x-1) ) -1/2 x/( x^2 +1) - 1/(2 (x^2 +1) )

Therefore

 ∫ x^4/( (x-1) (x^2 +x+1) ) dx = x^2/2 +x+1/2 log | x-1| -1/4 log (x^2 +1 ) -1/2 tan^(-1) x +C
Q 3116334270 Find ∫ [ log ( log x ) + 1/(( log x)^2) ] dx
Class 12 Chapter 7 Example 40 Solution:

Let I = ∫ [ log (log x) +1/(log x)^2 ] dx

= ∫ log (log x) dx + ∫ 1/(log x)^2 dx

In the first integral, let us take 1 as the second function. Then integrating it by
parts, we get

I = x log (log x ) - ∫ 1/(x log x ) dx + ∫ (dx)/(log x)^2

= x log (log x) - ∫ (dx)/(log x) + ∫ (dx)/( log x)^2 ...........(1)

Again, consider  ∫ (dx)/(log x)  , take 1 as the second function and integrate it by parts,

we have  ∫ (dx)/(log x) = [x/(log x) - ∫ x { -1/(log x)^2 (1/x) } dx] ......(2)

Putting (2) in (1), we get

I = x log ( log x ) -x/(log x) - ∫ (dx)/(log x)^2 + ∫ (dx)/(log x)^2 = x log ( log x) -x/(log x) +C
Q 3116123079 Find  ∫ [ log ( log x) + 1/(log x)^2] dx
Class 12 Chapter 7 Example 40 Solution:

Let I= ∫ [ log (logx ) + 1/(log x)^2 ] dx

= ∫ log (logx ) dx + ∫ 1/(log x)^2 dx

In the first integral, let us take 1 as the second function. Then integrating it by
parts, we get

I= x log ( log x) - ∫ 1/(x log x) x dx + ∫ (dx)/(log x)^2

= x log ( log x ) -∫ (dx)/(log x) + ∫ (dx)/( log x)^2 ......(1)

Again, consider  ∫ (dx)/(log x)  , take 1 as the second function and integrate it by parts,

we have  ∫ (dx)/( log x) = [ x/(log x) - ∫ x { 1/( log x)^2 (1/x) } dx ] .........(2)

Putting (2) in (1), we get

I = x log ( log x) -x/(log x) - ∫ (dx)/( (log x)^2) + ∫ (dx)/( ( log x)^2) = x log ( log x) - x/(log x) +C
Q 3136223172 Find  ∫ [ sqrt (cot x) + sqrt (tan x) ] dx
Class 12 Chapter 7 Example 41 Solution:

We have

I= ∫ [sqrt (cot x) + sqrt (tan x) ] dx = ∫ sqrt (tan x) (1+cot x) dx

Put tan x = t^2, so that sec^2 x dx = 2t dt

or dx= (2t dt)/( 1+t^4)

Then I= ∫ t (1+1/t^2) (2t)/(1+t^4) dt

= 2 ∫ (t^2 +1)/( t^4 +1) = dt = 2 ∫ ( (1+1/t^2 ) dt)/( t^2 +1/t^2 ) =2 ∫ ( ( 1+1/t^2 ) dt)/( (t-1/t)^2 +2)

put  t -1/t = y , so that  (1+1/t^2 ) dt = dy  . then

I = 2 ∫ (dy)/( y^2 + (sqrt 2)^2 ) = sqrt 2 tan^(-1) y/(sqrt 2) +C = sqrt 2 tan^(-1) ( t-1/t)/(sqrt 2) +C

= sqrt 2 tan^(-1) ( (t^2 -1)/( sqrt 2 t) ) +C = sqrt 2 tan^(-1) ( (tan x -1)/( sqrt (2 tan x) ) ) +C
Q 3146223173 Find  ∫ (sin 2x cos 2x dx )/( sqrt (9 - cos^4 (2x) ) )
Class 12 Chapter 7 Example 42 Solution:

Let I = ∫ (sin 2x cos 2x)/(sqrt ( 9- cos^4 2x) ) dx

Put cos^2 (2x) = t so that 4 sin 2x cos 2x dx = – dt

Therefore I = -1/4 ∫ (dt)/(sqrt (9- t^2) ) = -1/4 sin^(-1) (t/3) +C = -1/4 sin^(-1) [1/3 cos^2 2x] +C
Q 3156223174 Evaluate  ∫_(-1)^(3/2) | x sin ( pi x) | dx
Class 12 Chapter 7 Example 43 Solution:

Here f (x) = | x sin πx | = { tt ( (x sin pi x text (for ) -1 le x le 1 ), ( -x sin pi x text ( for ) 1 le x le 3/2 ) )

Therefore  ∫_(-1)^(3/2) | x sin pi x | dx = ∫_(-1)^1 x sin pi x dx + ∫_(1)^(3/2) - x sin pi x dx

= ∫_(-1)^1 x sin pi x dx - ∫_(1)^(3/2) x sin pi x dx

Integrating both integrals on righthand side, we get

 ∫_(-1)^(3/2) | x sin pi x| dx = [ (-x cos pi x)/pi + (sin pi x)/pi^2 ]_(-1)^1 - [ (-x cos pi x)/pi + ( sin pi x)/pi^2 ]_(1)^(3/2)

= 2/pi - [-1/pi^2 -1/pi] = 3/pi +1/pi^2
Q 3166223175 Evaluate ∫_0^pi (x dx)/(a^2 cos^2 x + b^2 sin^2 x)
Class 12 Chapter 7 Example 44 Solution:

Let I = ∫_0^pi (x dx)/( a^2 cos^2 x+ b^2 sin^2 x) = ∫_(0)^pi ( ( pi- x) dx)/(a^2 cos^2 (pi-x) +b^2 sin^2 (pi-x)) (using P_4)

= pi ∫_(0)^pi (dx)/( a^2 cos^2 x + b^2 sin^2 x) - ∫_0^pi (x dx)/( a^2 cos^2 x + b^2 sin^2 x)

= pi ∫_(0)^pi (dx)/( a^2 cos^2 x+ b^2 sin^2 x) -I

Thus 2I = pi ∫_0^pi (dx)/( a^2 cos^2 x + b^2 sin^2 x)

or I = pi/2 _0^pi (dx)/( a^2 cos^2 x+ b^2 sin^2 x) = pi/2 * 2 ∫_0^(pi/2) (dx)/( a^2 cos^2 x + b^2 sin^2 x)

(using P_6 )

= pi ∫_0^(pi/2) (sec^2 x dx)/( a^2 + b^2 tan^2 x) (dividing numerator and denominator by cos^2 x).

Put b tan x = t, so that b sec^2 x dx = dt. Also, when x = 0, t = 0, and when x = pi/2,  t -> oo

Therefore, I = pi/b ∫_0^oo (dt)/(a^2 + t^2) = pi/b * 1/a [ tan^(-1) (t/a)]_(0)^oo = pi/(ab)
[ pi/2 -0 ] = pi^2/(2 ab) 