# RD Sharma Class 12 Ex 19.24 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.24 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.24 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.24 Solutions Chapter 19 Indefinite Integrals

### Question 1. ∫dx/(1-cotx)

Solution:

We have,

Let I=∫dx/(1-cotx)

=∫sinx.dx/(sinx-cosx)

=(1/2)∫2sinx.dx/(sinx-cosx)

=(1/2)∫[(sinx+cosx)dx/(sinx-cosx)]+(1/2)∫dx

Let, sinx-cosx=z

Differentiating both sides we have

(cosx+sinx)dx=dz

=(1/2)Log|sinx-cosx|+(x/2)+C  (Here C is integration constant)

### Question 2. ∫dx/(1-tanx)

Solution:

We have,

Let I=∫dx/(1-tanx)

=∫cosx.dx/(cosx-sinx)

=(1/2)∫2cosx.dx/(cosx-sinx)

=(1/2)∫[(cosx+sinx)dx/(cosx-sinx)]+(1/2)∫dx

Let, cosx-sinx=z

Differentiating both sides we have

-(sinx+cosx)dx=dz

(sinx+cosx)dx=-dz

=(x/2)-(1/2)Log|cosx-sinx|+C  (Here C is integration constant)

### Question 3. ∫[(3+2cosx+4sinx)/(2sinx+cosx+3)]dx

Solution:

We have,

Let I=∫[(3+2cosx+4sinx)/(2sinx+cosx+3)]dx

Now substituting numerator

3+2cosx+4sinx=A(d/dx)(2sinx+cosx+3)+B(2sinx+cosx+3)+C

3+2cosx+4sinx=A(2cosx-sinx)+B(2sinx+cosx+3)+C

3+2cosx+4sinx=2Acosx-Asinx+2Bsinx+Bcosx+3B+C

3+2cosx+4sinx=(3B+C)+(2A+B)cosx+(2B-A)sinx

(3B+C)=3                         (i)

(2A+B)=2                        (ii)

(2B-A)=4                        (iii)

On solving above equations,

A=0 ,B=2 ,C=-3

=2∫dx-∫3/(2sinx+cosx+3)

=I1-I2

I1=2∫dx

=2x

I2=∫3/(2sinx+cosx+3)

Substituting  and

Let, tan(x/2)=z

Differentiating both sides,

(1/2)sec2(x/2)dx=dz

=3∫dz/(z2+2z+2)

=3∫dz/(z2+2z+1+1)

=3∫dz/{(z+1)2+12}

=3tan-1(z+1)

Putting the value of z

=3tan-1(tanx/2+1)

I2=3tan-1(tanx/2+1)

I=I1-I2

=2x-3tan-1(tanx/2+1)+C    (Here C is integration constant)

### Question 4. ∫dx/(p+qtanx)

Solution:

We have,

Let I=∫dx/(p+qtanx)

=∫[cosx/(pcosx+qsinx)]dx

Now substituting numerator

cosx=A(d/dx)(pcosx+qsinx)+B(pcosx+qsinx)+C

cosx=A(-psinx+qcosx)+B(pcosx+qsinx)+C

cosx=sinx(Bq-Ap)+cosx(Bp+Aq)+C

On comparing both sides,

C=0,

Bp+Aq=1,

Bq-Ap=0,

Solving above equation,

A=q/(p2+q2) and B=p/(p2+q2

[Tex]I=∫\frac{\frac{q}{p^2+q^2}(-psinx+qcosx)+\frac{p}{p^2+q^2}(pcosx+qsinx)}{pcosx+qsinx}dx[/Tex]

I=I1+I2

I1=

=q/(p2+q2)log|pcosx+qsinx|

I2=

=px/(p2+q2)

I=q/(p2+q2)log|pcosx+qsinx|+px/(p2+q2)+C     (Here C is integration constant)

### Question 5. ∫[(5cosx+6)/(2cosx+sinx+3)]dx

Solution:

We have,

Let I=∫[(5cosx+6)/(2cosx+sinx+3)]dx

Now substituting numerator

5cosx+6=A(d/dx)(2cosx+sinx+3)+B(2cosx+sinx+3)+C

5cosx+6=A(-2sinx+cosx)+B(2cosx+sinx+3)+C

5cosx+6=sinx(B-2A)+cosx(2B+A)+3B+C

On comparing both sides,

3B+C=6,

2B+A=5,

B-2A=0,

Solving above equation,

A=1, B=2 and c=0

I=I1+I2

I1=∫[(-2sinx+cosx)/(2cosx+sinx+3)]dx

I1=log|2cosx+sinx+3|

I2=2∫dx

I2=2x

I=log|2cosx+sinx+3|+2x+C      (Here C is integration constant)

### Question 6. ∫[(2sinx+3cosx)/(3sinx+4cosx)]dx

Solution:

We have,

Let I=∫[(2sinx+3cosx)/(3sinx+4cosx)]dx

Now substituting numerator

2sinx+3cosx=A(d/dx)(3sinx+4cosx)+B(3sinx+4cosx)+C

2sinx+3cosx=A(3cosx-4sinx)+B(3sinx+4cosx)+C

2sinx+3cosx=sinx(3B-4A)+cosx(4B+3A)+3B+C

On comparing both sides,

3B-4A=2,

4B+3A=3,

Solving above equation,

A=1/25, B=18/25 and C=0

I=I1+I2

I1=(1/25)∫[(3cosx-4sinx)/(3sinx+4cosx)]dx

I1=(1/25)log|3sinx+4cosx|

I2=(18/25)∫dx

I2=(18x/25)

I=(1/25)log|3sinx+4cosx|+(18x/25)+C      (Here C is integration constant)

### Question 7. ∫dx/(3+4cotx)

Solution:

We have,

Let I=∫dx/(3+4cotx)

=∫[(sinx)/(3sinx+4cosx)]dx

Now substituting numerator

sinx=A(d/dx)(3sinx+4cosx)+B(3sinx+4cosx)+C

sinx=A(3cosx-4sinx)+B(3sinx+4cosx)+C

sinx=sinx(3B-4A)+cosx(4B+3A)+3B+C

On comparing both sides,

3B-4A=1,

4B+3A=0,

Solving above equation,

A=-4/25, B=3/25 and C=0

I=I1+I2

I1=(-4/25)∫[(3cosx-4sinx)/(3sinx+4cosx)]dx

I1=(-4/25)log|3sinx+4cosx|

I2=(3/25)∫dx

I2=(3x/25)

I=(3x/25)-(4/25)log|3sinx+4cosx|+(18x/25)+C      (Here C is integration constant)

### Question 8. ∫[(2tanx+3)/(3tanx+4)]dx

Solution:

We have,

Let I=∫[(2tanx+3)/(3tanx+4)]dx

=∫[(2sinx+3cosx)/(3sinx+4cosx)]dx

Now substituting numerator

2sinx+3cosx=A(d/dx)(3sinx+4cosx)+B(3sinx+4cosx)+C

2sinx+3cosx=A(3cosx-4sinx)+B(3sinx+4cosx)+C

2sinx+3cosx=sinx(3B-4A)+cosx(4B+3A)+C

On comparing both sides,

3B-4A=2,

4B+3A=3,

Solving above equation,

A=1/25, B=18/25 and C=0

I=I1+I2

I1=(1/25)∫[(3cosx-4sinx)/(3sinx+4cosx)]dx

I1=(1/25)log|3sinx+4cosx|

I2=(18/25)∫dx

I2=(18x/25)

I=(1/25)log|3sinx+4cosx|+(18x/25)+C      (Here C is integration constant)

### Question 9. ∫dx/(4+3tanx)

Solution:

We have,

Let I=∫dx/(4+3tanx)

=∫[(cosx)/(4cosx+3sinx)]dx

Now substituting numerator

cosx=A(d/dx)(3sinx+4cosx)+B(3sinx+4cosx)+C

cosx=A(3cosx-4sinx)+B(3sinx+4cosx)+C

cosx=sinx(3B-4A)+cosx(4B+3A)+C

On comparing both sides,

3B-4A=0,

4B+3A=3,

Solving above equation,

A=3/25, B=4/25 and C=0

I=I1+I2

I1=(3/25)∫[(3cosx-4sinx)/(3sinx+4cosx)]dx

I1=(3/25)log|3sinx+4cosx|

I2=(4/25)∫dx

I2=(4x/25)

I=(3/25)log|3sinx+4cosx|+(4x/25)+C      (Here C is integration constant)

### Question 10. ∫[(8cotx+1)/(3cotx+2)]dx

Solution:

We have,

Let I=∫[(8cotx+1)/(3cotx+2)]dx

=∫[(8cosx+sinx)/(3cosx+2sinx)]dx

Now substituting numerator

8cosx+sinx=A(d/dx)(3cosx+2sinx)+B(3cosx+2sinx)+C

8cosx+sinx=A(-3sinx+2cosx)+B(3cosx+2sinx)+C

8cosx+sinx=sinx(2B-3A)+cosx(3B+2A)+C

On comparing both sides,

2B-3A=1,

3B+2A=3,

Solving the above equation,

A=1, B=2 and C=0

I=I1+I2

I1=∫[(-3sinx+2cosx)/(3cosx+2sinx)]dx

I1=log|3cosx+2sinx|

I2=2∫dx

I2=2x

I=log|3cosx+2sinx|+2x+C      (Here C is integration constant)

### Question 11. ∫[(4sinx+5cosx)/(5sinx+4cosx)]dx

Solution:

We have,

Let I=∫[(4sinx+5cosx)/(5sinx+4cosx)]dx

Now substituting numerator

4sinx+5cosx=A(d/dx)(5sinx+4cosx)+B(5sinx+4cosx)+C

4sinx+5cosx=A(5cosx-4sinx)+B(5sinx+4cosx)+C

4sinx+5cosx=sinx(5B-4A)+cosx(4B+5A)+C

On comparing both sides,

5B-4A=4,

4B+5A=5,

Solving the above equation,

A=9/41, B=40/41 and C=0

I=I1+I2

I1=(9/41)∫[(5cosx-4sinx)/(5sinx+4cosx)]dx

I1=(9/41)log|5sinx+4cosx|

I2=(40/41)∫dx

I2=(40x/41)

I=(9/41)log|5sinx+4cosx|+(40x/41)+C      (Here C is integration constant)

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