Mathematics Some Properties of Definite Integrals For CBSE-NCERT
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Class 12 Chapter 7 - Integrals

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)`
 
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