Mathematics Integration using trigonometric identities and Integrals of Some Particular Functions FOR CBSE-NCERT
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Class 12 Chapter 7 - Integrals

Integration using trigonometric identities and Integrals of Some Particular Functions FOR CBSE-NCERT

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`color{red} {star}` Integration using trigonometric identities
`color{red} star` Integrals of Some Particular Functions

Integration using trigonometric identities

● When the integration involves some trigonometric functions, we use some known identities to find the integral as illustrated through the following example.
Q 3135478362

Find

(i) `∫cos^2x dx`

(ii) `∫sin 2x cos 3x dx`

(iii) `∫sin^3 x dx`


Class 12 Chapter 7 Example 7
Solution:

(i) Recall the identity `cos 2x = 2 cos^2 x – 1`, which gives

`cos^2 x = (1+ cos 2x)/2`

Therefore, `∫ cos^2 x dx =1/2 ∫ (1+ cos 2x) dx = 1/2 ∫ dx +1/2 ∫ cos 2x dx`

`= x/2 +1/4 sin 2x +C`


(ii) Recall the identity `sin x cos y =1/2 [sin (x + y) + sin (x – y)]` (Why?)

Then `∫ sin 2x cos 3x dx = 1/2 [∫ sin 5x dx • ∫ sin x dx ]`

`=1/2 [ -1/5 cos 5x + cos x ] +C`

`= -1/10 cos 5x + 1/2 cos x +C`


(iii) From the identity `sin 3x = 3 sin x – 4 sin^3 x` , we find that

` sin^3 x = (3 sin x - sin 3x)/4`

Therefore,` ∫ sin^3 x dx = 3/4 ∫ sin x dx - 1/4 ∫ sin 3x dx`

` = -3/4 cos x + 1/12 cos 3x +C`

Alternatively, `∫ sin^3 x dx = ∫ sin^2 x sin x dx = ∫(1 – cos^2 x) sin x dx`

Put `cos x = t` so that` – sin x dx = dt`

Therefore, ` ∫ sin^3 x dx = - ∫ (1-t^2) dt = - ∫ dt + ∫ t^2 dt = - t + t^3/3 +C`

`= -cos x +1/3 cos^3 x +C`

Integrals of Some Particular Functions

Some important formulae of integrals

(1) `color{green}{ ∫ (dx)/(x^2 -a^2) =1/(2a) log | (x-a)/(x+a) | +C}`

(2) `color{green}{∫ (dx)/(a^2 - x^2) = 1/(2a) log | (a+x)/(a-x) | +C}`

(3 ) ` color{green}{∫ (dx)/(x^2 +a^2) =1/a tan^(-1) (x/a) +C}`

(4) `color{green}{∫ (dx)/( sqrt (x^2-a^2) ) = log | x+ sqrt ( x^2 - a^2) | +C}`

(5) `color{green}{∫ (dx)/(sqrt (a^2-x^2) ) = sin^(-1) (x/a) +C}`

(6) `color{green}{∫ (dx)/(sqrt (x^2 + a^2 ) ) = log | x+ sqrt ( x^2 +a^2 ) | +C}`


We now prove the above results:

(1) We have `color{orange}{1/(x^2 -a^2 ) = 1/( (x-a)(x+a) )}` (key line)

`color{blue} {= 1/(2a) [ ( (x+a)- (x-a) )/( (x-a) (x+a) ) ] = 1/(2a) [1/(x-a) -1/(x+a) ]}`

Therefore, `color{blue} { ∫ (dx)/(x^2 -a^2) = 1/(2a) [ ∫ (dx)/(x-a) - ∫ (dx)/(x+a) ]}`

`color{blue} {=1/(2a) [ log| (x-a) | -log | (x-a) | ] +C}`

`color{blue} { = 1/(2a) log | (x-a)/(x+a) | +C}`

(2) In view of (1) above, we have

`color{orange}{1/(a^2 - x^2 ) = 1/( (a-x)(a+x) )}` (key line)

`color{blue} {1/(2a) [ ( (a+x) + (a-x) )/( (a+x)( a-x) ) ] =1/(2a) [1/(a-x) + 1/(a+x) ]}`

Therefore, ` color{blue} {∫ (dx)/(a^2 - x^2) = 1/(2a) [ ∫ (dx)/(a-x) + ∫ (dx)/(a+x) ]}`

`color{blue} {=1/(2a) [ - log | a-x} + log |a +x | ] +C}`

` color{blue} {= 1/(2a) log | (a+x)/(a-x) | +C}`.

(3) `color{orange}{"Put" x = a tan θ.}` Then `dx = a sec^2 θ dθ`.

Therefore, ` color{blue} {∫ (dx)/( x^2 +a^2 ) = ∫ (a sec^2 theta d theta )/(a^2 tan^2 theta +a^2)}`

`color{blue} {=1/a ∫ d theta = 1/a theta +C = 1/a tan^(-1) (x/a) +C}`

(4) `color{orange}{"Let "\ \ x = a secθ.}` Then dx = a secθ tan θ dθ.

Therefore, ` color{blue} {∫ (dx)/( sqrt (x^2- a^2) ) = ∫ (a sec theta tan theta d theta )/( sqrt (a^2 sec^2 theta- a^2) )}`

` color{blue} {= ∫secθ dθ = log | sec θ + tan θ| + C_1}`

`color{blue} {= log | x/a + sqrt ( x^2/a^2 -1 ) | +C_1}`

` color{blue} {= log | x+ sqrt (x^2 -a^2 ) | -log |a| + C_1}`

`color{blue} {= log | x + sqrt (x^2 - a^2 ) | +C` , where `C =C_1 = log |a|}`


(5) `color{orange}{"Let" \ \ x = a sinθ.}` Then dx = a cosθ dθ.

Therefore, ` color{blue} {∫ (dx)/(sqrt (a^2 - x^2 ) ) = ∫ (a cos theta d theta )/(sqrt (a^2 -a^2 sin^2 theta) )}`

`color{blue} {= ∫ d theta = theta +C = sin^(-1) (x/a) +C}`


(6) `color{orange}{"Let"\ \ x = a tanθ.}' Then ``dx = a sec^2 θ dθ`.

Therefore, ` color{blue} {∫ (dx)/( sqrt (x^2 +a^2) ) = ∫ (a sec^2 theta d theta )/(sqrt (a^2 tan^2 theta + a^2 ) )}`

`color{blue} {= ∫secθ dθ = log (secθ + tanθ) + C_1}`

`color{blue} {= log | x/a + sqrt (x^2/a^2 +1 ) | +C_1}`

`color{blue} {= log | x + sqrt (x^2 +a^2) | - log |a | +C_1}`

`color{blue} {= log | x + sqrt (x^2 +a^2) | +C` , where `C= C_1 -log |a|}`


Useful techniques directly use to evaluate integrals

(1)● To find the integral `color{red}{ ∫ (dx)/( ax^2 + bx +c)}` , we write

`=>`write it as ` color{green} {ax^2 + bx + c =a [ x^2 + b/a x + c/a ] }`

`= color{green} {a [(x+ b/(2a) )^2 + (c/a - b^2/(4a^2)) ]}`

`=>` Now, put `x + b/(2a) = t` so that dx = dt and writing ` color{blue} {c/a - b^2/(4a^2) = pm k^2}` We find the

integral reduced to the form `1/a ∫ (dt)/( t^2 pm k^2)` depending upon the sign of ` (c/a - b^2/(4a^2 ))` and hence can be evaluated.

(2)● To find the integral of the type `color{red}{ ∫ (dx)/sqrt(ax^2 + bx +c)}` , proceeding as in (1).

(3)● To find the integral of the type ` color{red}{∫ (px +q)/(ax^2 +bx +c)dx}` , where p, q, a, b, c are constants, we are to find real numbers A, B such that

`=>` `color {green }{px + q = A d/(dx) (ax^2 + bx+c) +B = A (2ax + b) +B}`

`=>` To determine A and B, we equate from both sides the coefficients of x and the constant terms. A and B are thus obtained and hence the integral is reduced to one of the known forms.

(4) ● For the evaluation of the integral of the type `color{green}{ ∫ ( (px+q)dx)/( sqrt (ax^2 +bx +c) ) }` , we proceed

as in (3) and transform the integral into known standard forms.


Let us illustrate the above methods by some examples.
Q 3155478364

Find the following integrals:

(i) ` ∫ (dx)/(x^2 -16)`

(ii) `∫ (dx)/( sqrt (2x -x^2 ) )`
Class 12 Chapter 7 Example 8
Solution:

(i) We have ` ∫ (dx)/(x^2 -16) = ∫ (dx)/( x^2 - 4^2) = 1/8 log | (x-4)/(x+4) | +C` [by 7.4 (1)]

(ii) ` ∫ (dx)/( sqrt (2x - x^2) ) = ∫ (dx)/(sqrt ( 1- (x-1)^2 ) )`

Put x – 1 = t. Then dx = dt.

Therefore, ` ∫ (dx)/( sqrt (2x -x^2 ) ) = ∫ (dt)/(sqrt (1-t^2 ) ) = sin^(-1) (t) +C`[by 7.4 (5)]

`= sin^(-1) (x-1) +C`
Q 3115478369

Find the following integrals :

(i) ` ∫ (dx)/( x^2 - 6x +13)`

(ii) ` ∫ (dx)/( 3x^2 +13x -10)`

(iii) ` ∫ (dx)/(sqrt (5x^2 -2x) )`
Class 12 Chapter 7 Example 9
Solution:

(i) We have `x^2 – 6x + 13 = x^2 – 6x + 3^2 – 3^2 + 13 = (x – 3)^2 + 4`

So, ` ∫ (dx)/(x^2 -6x +13) = ∫ 1/( (x-3)^2 +2^2 ) dx`

Let `x-3 = t` , Then dx = dt

Therefore, ` ∫ (dx)/(x^2 -6x +13) = ∫ (dt)/(t^2 +2^2) =1/2 tan^(-1) (t/2) +C` [by 7.4 (3)]

` =1/2 tan^(-1) ((x-3)/2 ) +C`

(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,

` 3x^2 +13x – 10 = 3 (x^2 + (13x)/3 -10/3)`

`= 3 [ (x +13/6 )^2 - (17/6)^2 ]` (completing the square)


Thus ` ∫ (dx)/( 3x^2 +13x -10) = 1/3 ∫ (dx)/( (x+13/6)^2 - (17/6)^2 )`

Put `x +13/6 =t` .Then `dx = dt`


Therefore, `int (dx)/(3x^2 +13x -10) = 1/3 ∫ (dt)/( t^2 - (17/6)^2 )`


` = 1/(3 xx 2 xx 17/6) log | (t-17/6)/(t +17/6) | +C_1` [by 7.4 (i)]

` = 1/17 log | ( x+13/6 -17/6)/( x+ 13/6 +17/6) | +C_1`

` = 1/17 log | (6x-4)/(6x+30) | +C_1`


` = 1/17 log | (3x-2)/(x+5) | +C_1 + 1/17 log 1/3`

` = 1/17 log | (3x-2)/(x+5) | +C` , where `C= C_1 +1/17 log 1/3`

(iii) We have ` ∫ (dx)/( sqrt (5x^2 -2x) ) = ∫ (dx)/( sqrt ( 5 (x^2 - (2x)/5) ) )`

`= 1/(sqrt 5) ∫ (dx)/( sqrt ( ( x-1/5)^2 - (1/5)^2 ) )` (completing the square)

Put `x - 1/5 = t ` . Then `dx = dt`

Therefore, ` ∫ (dx)/(sqrt (5x^2 -2x) ) = 1/(sqrt 5) ∫ (dt)/( sqrt (t^2 - (1/5)^2 ) )`

`= 1/(sqrt 5) log | t + sqrt (t^2 - (1/5)^2 ) | +C` [by 7.4 (4)]

` = 1/(sqrt 5) log | x-1/5 + sqrt ( x^2 - (2x)/5) | +C`
Q 3185578467

Find the following integrals:

(i) ` ∫ (x+2)/(2x^2 + 6x +5) dx`

(ii) ` ∫ (x+3)/(sqrt (5-4 x + x^2) ) dx`

Class 12 Chapter 7 Example 10
Solution:

(i) Using the formula 7.4 (9), we express

`x+ 2 = A d/(dx) (2x^2 + 6x + 5 ) +B =A (4x + 6) +B`

Equating the coefficients of x and the constant terms from both sides, we get

`4A =1` and ` 6A+ B =2` or `A =1/4 ` and `B = 1/2` .

Therefore, ` ∫ (x+2)/(2x^2 + 6x +5) =1/4 ∫ (4x+6)/(2x^2 + 6x +5) dx +1/2 ∫ (dx)/(2x^2 + 6x +5 )`

` =1/4 I_1 + 1/2 I_2` (say) ... (1)

In `I_1`, put `2x^2 + 6x + 5 = t`, so that `(4x + 6) dx = dt`

Therefore, `I_1 = ∫ (dt)/t = log | t | +C_1`

`= log | 2x^2 + 6x +5 | + C_1` .........(2)

and `I_2 = ∫ (dx)/(2x^2 + 6x + 5) = 1/2 ∫ (dx)/( x^2 +3x + 5/2)`

`=1/2 ∫ (dx)/( (x+3/2)^2 + (1/2)^2 )`

Put `x +3/2 = t` , so that dx = dt, we get

`I_2 = 1/2 ∫ (dt)/( t^2 + (1/2)^2 ) =1/(2 xx 1/2) tan^(-1) 2t +C_2` [by 7.4 (3)]

`= tan^(-1) 2 (x +3/2) +C_2 = tan^(-1) (2x+3) +C_2` ..............(3)

Using (2) and (3) in (1), we get

` ∫ (x+2)/(2x^2 + 6x +5) dx = 1/4 log |2x^2 + 6x + 5 | +1/2 tan^(-1) (2x +3) +C`

where, `C= (C_1)/4 + (C_2)/2`

(ii) This integral is of the form given in 7.4 (10). Let us express

` x+3 = A d/(dx) (5- 4x - x^2 ) +B =A (-4 -2x) +B`

Equating the coefficients of x and the constant terms from both sides, we get

`– 2A = 1` and `– 4 A + B = 3`, i.e., `A =-1/2 ` and `B=1`

Therefore, ` ∫ (x+3)/( sqrt (5- 4x - x^2) ) dx = -1/2 ∫ ( (-4-2x) dx)/( sqrt (5-4 x - x^2) ) +∫ (dx)/( sqrt (5- 4x - x^2 ) )`

` = -1/2 I_1 +I_2` ....... (1)

In `I_1`, put `5 – 4x – x^2 = t`, so that `(– 4 – 2x) dx = dt`.

Therefore, `I_1 = ∫ ( ( -4-2x) dx)/( sqrt (5-4x -x^2) ) = ∫ (dt)/(sqrt t ) = 2 sqrt t +C_1`

` = 2 sqrt (5-4 x - x^2) +C_1` ...... (2)

Now consider ` I_2 = ∫ (dx)/( sqrt (5-4x - x^2) ) = ∫ (dx)/( sqrt (9- (x+2)^2 ) )`

Put x + 2 = t, so that dx = dt.

Therefore, ` I_2 = ∫ (dt)/(sqrt (3^2 - t^2 ) ) = sin^(-1) (t/3) +C_2` [by 7.4 (5)]

`= sin^(-1) ((x+2)/3) +C_2` ....... (3)


Substituting (2) and (3) in (1), we obtain

` ∫ (x+3)/(sqrt (5-4x - x^2) ) = - sqrt (5- 4x -x^2) + sin^(-1) ((x+2)/3) +C` , where `C = C_2 - C_1/2`
 
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