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

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)