 Mathematics 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 