# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.24

### Question 1. âˆ«dx/(1-cotx)

Solution:

We have,

Let I=âˆ«dx/(1-cotx)

=âˆ«sinx.dx/(sinx-cosx)

=(1/2)âˆ«2sinx.dx/(sinx-cosx)

=(1/2)âˆ«[(sinx+cosx)dx/(sinx-cosx)]+(1/2)âˆ«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. âˆ«dx/(1-tanx)

Solution:

We have,

Let I=âˆ«dx/(1-tanx)

=âˆ«cosx.dx/(cosx-sinx)

=(1/2)âˆ«2cosx.dx/(cosx-sinx)

=(1/2)âˆ«[(cosx+sinx)dx/(cosx-sinx)]+(1/2)âˆ«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. âˆ«[(3+2cosx+4sinx)/(2sinx+cosx+3)]dx

Solution:

We have,

Let I=âˆ«[(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

=2âˆ«dx-âˆ«3/(2sinx+cosx+3)

=I1-I2

I1=2âˆ«dx

=2x

I2=âˆ«3/(2sinx+cosx+3)

Substituting  and

Let, tan(x/2)=z

Differentiating both sides,

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

=3âˆ«dz/(z2+2z+2)

=3âˆ«dz/(z2+2z+1+1)

=3âˆ«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. âˆ«dx/(p+qtanx)

Solution:

We have,

Let I=âˆ«dx/(p+qtanx)

=âˆ«[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=âˆ«\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. âˆ«[(5cosx+6)/(2cosx+sinx+3)]dx

Solution:

We have,

Let I=âˆ«[(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=âˆ«[(-2sinx+cosx)/(2cosx+sinx+3)]dx

I1=log|2cosx+sinx+3|

I2=2âˆ«dx

I2=2x

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

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

Solution:

We have,

Let I=âˆ«[(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)âˆ«[(3cosx-4sinx)/(3sinx+4cosx)]dx

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

I2=(18/25)âˆ«dx

I2=(18x/25)

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

### Question 7. âˆ«dx/(3+4cotx)

Solution:

We have,

Let I=âˆ«dx/(3+4cotx)

=âˆ«[(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)âˆ«[(3cosx-4sinx)/(3sinx+4cosx)]dx

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

I2=(3/25)âˆ«dx

I2=(3x/25)

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

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

Solution:

We have,

Let I=âˆ«[(2tanx+3)/(3tanx+4)]dx

=âˆ«[(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)âˆ«[(3cosx-4sinx)/(3sinx+4cosx)]dx

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

I2=(18/25)âˆ«dx

I2=(18x/25)

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

### Question 9. âˆ«dx/(4+3tanx)

Solution:

We have,

Let I=âˆ«dx/(4+3tanx)

=âˆ«[(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)âˆ«[(3cosx-4sinx)/(3sinx+4cosx)]dx

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

I2=(4/25)âˆ«dx

I2=(4x/25)

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

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

Solution:

We have,

Let I=âˆ«[(8cotx+1)/(3cotx+2)]dx

=âˆ«[(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=âˆ«[(-3sinx+2cosx)/(3cosx+2sinx)]dx

I1=log|3cosx+2sinx|

I2=2âˆ«dx

I2=2x

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

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

Solution:

We have,

Let I=âˆ«[(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)âˆ«[(5cosx-4sinx)/(5sinx+4cosx)]dx

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

I2=(40/41)âˆ«dx

I2=(40x/41)

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

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