RD Sharma Class 12 Ex 19.24 Solutions Chapter 19 Indefinite Integrals

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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.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.24
CategoryRD Sharma Solutions

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)

=∫\frac{dx}{1-\frac{cosx}{sinx}}

=∫sinx.dx/(sinx-cosx)

=(1/2)∫2sinx.dx/(sinx-cosx)

=\frac{1}{2}∫\frac{sinx+cosx+sinx-cosx}{sinx-cosx}dx

=(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)

=∫\frac{dx}{1-\frac{sinx}{cosx}}

=∫cosx.dx/(cosx-sinx)

=(1/2)∫2cosx.dx/(cosx-sinx)

=\frac{1}{2}∫\frac{cosx+sinx+cosx-sinx}{cosx-sinx}dx

=(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

I=∫\frac{2(2sinx+cosx+3)-3}{2sinx+cosx+3)}

=2∫dx-∫3/(2sinx+cosx+3)

=I1-I2

I1=2∫dx

=2x

I2=∫3/(2sinx+cosx+3)

Substituting cosx=\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}  and sinx=\frac{2tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}

=3∫\frac{dx}{2×\frac{2tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}+\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}+3}
=3∫\frac{(1+tan^2\frac{x}{2})}{4tan\frac{x}{2}+1-tan^2\frac{x}{2}+3(1+tan^2\frac{x}{2})}
=\frac{3}{2}∫\frac{sec^2\frac{x}{2}}{tan^2\frac{x}{2}+2tan\frac{x}{2}+2}

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) 

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

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

I2=∫\frac{\frac{p}{p^2+q^2}(pcosx+qsinx)}{pcosx+qsinx}dx

=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=∫\frac{(-2sinx+cosx)+2(2cosx+sinx+3)}{2cosx+sinx+3}dx

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=∫\frac{\frac{1}{25}(3cosx-4sinx)+\frac{18}{25}(3sinx+4cosx)}{(3sinx+4cosx)}dx

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=∫\frac{\frac{-4}{25}(3cosx-4sinx)+\frac{3}{25}(3sinx+4cosx)}{(3sinx+4cosx)}dx

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=∫\frac{\frac{1}{25}(3cosx-4sinx)+\frac{18}{25}(3sinx+4cosx)}{(3sinx+4cosx)}dx

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=∫\frac{\frac{3}{25}(3cosx-4sinx)+\frac{4}{25}(3sinx+4cosx)}{(3sinx+4cosx)}dx

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=∫\frac{(-3sinx+2cosx)+2(3cosx+2sinx)}{(3cosx+2sinx)}dx

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=∫\frac{\frac{9}{41}(5cosx-4sinx)+\frac{40}{41}(5sinx+4cosx)}{(5sinx+4cosx)}dx

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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