RD Sharma Class 12 Ex 19.16 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.16 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 19.16 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.16
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 19.16 Solutions Chapter 19 Indefinite Integrals

Question 1. Evaluate ∫ sec2x/ 1 – tan2x dx

Solution:

Let us assume I = ∫ sec2x/ 1 – tan2x dx              …..(i)

Now, put tan x = t

sec2x dx = dt

So, put all these values in eq(i)

= ∫ dt/ 1– t2 

On integrating the above equation then, we get

= 1/ 2(1) log|1 + t/1 – t| + c            

Since, ∫ 1/ a– x2 dx = 1/ 2a log|a + x/a – x| + c]

Hence, I = 1/2 log|1 + tanx/1 – tanx| + c

Question 2. Evaluate ∫ ex/ 1 + e2x dx

Solution:

Let us assume I = ∫ ex/ 1 + e2x dx              …..(i)

Now, put ex = t

ex dx = dt

So, put all these values in eq(i)

= dt/ 1 + t2 

On integrating the above equation then, we get

= tan-1t + c                     

Since, ∫ 1/ 1 + x2dx = tan-1x + c

Hence, I = tan-1ex + c 

Question 3. Evaluate ∫ cosx/ sin2x + 4sinx + 5 dx

Solution:

Let us assume I = ∫ cosx/ sin2x + 4sinx + 5 dx               …..(i)

Now, put sinx = t

cosx dx = dt

So, put all these values in eq(i)

= ∫ dt/ t+ 4t + 5

= ∫ dt/ t+ 2t(2) + (2)– (2)+ 5

= ∫ dt/ (t + 2)+ 1                      …..(ii)

Again, Put t + 2 = u

dt = du

Now, put all these values in eq(ii)

= ∫ du/ u+ 1

On integrating the above equation then, we get

= tan-1u + c              

Since, ∫1/ x+ 1 dx = tan-1x + c

= tan-1(t + 2) + c

Hence, I = tan-1(sinx + 2) + c

Question 4. Evaluate ∫ ex/e2x + 5e+ 6 dx

Solution:

Let us assume I = ∫ ex/e2x + 5e+ 6 dx                  …..(i)

Now, put ex = t

ex dx = dt

So, put all these values in eq(i)

= ∫ dt/ t+ 5t + 6

= ∫ dt/ t + 2t(5/2) + (5/2)– (5/2)+ 6

= ∫ dt/ (t + 5/2)– 1/4                      …..(ii)

Put t + 5/2 = u

dt = du

Now, put the above value in eq(ii)

= ∫ du/ u– (1/2)2 

On integrating the above equation then, we get

= 2/2 log|u – (1/2)/u + (1/2)| + c           

Since, ∫ 1/ x2 – a2 dx = 1/ 2alog|x – a/x + a| + c

= log|2u – 1/2u + 1| + c

= log|2(t + 5/2) – 1/2(t + 5/2) + 1| + c

Hence, I = log|e+ 2/e+ 3| + c

Question 5. Evaluate ∫ e3x/ 4e6x – 9 dx

Solution:

Let us assume I = ∫ e3x/ 4e6x– 9 dx                …..(i)

Now, put e3x = t

3e3x dx = dt

e3x dx = dt/3

Now, put the above value in eq(i)

= 1/3 ∫ dt/ 4t– 9

= 1/12 ∫ dt/ t– (3/2)2 

On integrating the above equation then, we get

= 1/12 x 1/ 2(3/2) log|t – 3/2/t + 3/2| + c        

Since, ∫1/ x– a2 dx = 1/2a log|x – a/x + a| + c]

= 1/36 log|2t – 3/2t + 3| + c

Hence, I = 1/36 log|2e3x – 3/2e3x + 3| + c

Question 6. Evaluate ∫ dx/e+ e-x

Solution:

Let us assume I = ∫ dx/e+ e-x  

=  dx/e+ 1/ex  

= ∫ exdx/ (ex)2 + 1                   …..(i)

Now, put ex = t

exdx = dt

Now, put the above value in eq(i)

= ∫ dt/ t+ 1

On integrating the above equation then, we get

= tan-1t + c                          

Since ∫ 1/ 1 + x2 dx = tan-1x + c

Hence, I = tan-1(ex) + c

Question 7. Evaluate ∫ x/ x+ 2x+ 3 dx

Solution:

Let us assume I = ∫ x/ x+ 2x+ 3 dx               …..(i)

Now, put x2 = t

2x dx = dt

x dx = dt/2

Now, put the above value in eq(i)

= 1/2 ∫ dt/ t+ 2t + 3

= 1/2 ∫ dt/ t+ 2t + 1 – 1 + 3

= 1/2 ∫ dt/ (t + 1)2 + 2                         …..(ii)

Now put t + 1 = u

dt = du

So, put the above value in eq(ii)

= 1/2 ∫ du/ u+ (√2)2

On integrating the above equation then, we get

= 1/2 x 1/√2 tan-1(u/√2) + c                  

Since ∫1/ x+ a2dx = 1/a tan-1(x/a) + c

= 1/2√2 tan-1(t + 1/ √2) + c

Hence, I = 1/2√2 tan-1(x+ 1/ √2) + c

Question 8. Evaluate ∫ 3x5/ 1 + x12 dx

Solution:

Let us assume I = ∫ 3x5/ 1 + x12 dx

= ∫ 3x5/ 1 + (x6)2dx                 …..(i)

Now, put x= t

6x5dx = dt

x5dx = dt/6

Now, put the above value in eq(i)

= 3/6 ∫ dt/ 1 + t2

On integrating the above equation then, we get

= 1/2 tan-1(t) + c                         

Since ∫ 1/ x+ 1 dx = tan-1x + c

Hence, I = 1/2 tan-1(x6) + c

Question 9. Evaluate ∫ x2/ x– a6 dx

Solution:

Let us assume I = ∫ x2/ x– a6 dx

= ∫ x2/ (x3)– (a3)2 dx                …..(i)

Now, put x3 = t

3x2 dx = dt

x2 dx = dt/3

Now, put the above value in eq(i)

= 1/3 ∫ dt/ t– (a3)2 

On integrating the above equation then, we get

= 1/3 x 1/2a3 log|t – a3/t + a3| + c                        

Since ∫1/ x– a2 dx = 1/2a log|x – a/x + a| + c

= 1/6a3 log|x– a3/x+ a3| + c

Hence, I = 1/6a3 log|x– a3/x+ a3| + c

Question 10. Evaluate ∫ x2/ x+ a6 dx

Solution:

Let us assume I = ∫ x2/ x+ a6 dx

= ∫ x2/ (x3)+ (a3)2 dx                 …..(i)

Now, put x3 = t

3x2 dx = dt

x2 dx = dt/3

Now, put the above value in eq(i)

= 1/3 ∫ dt/ t+ (a3)2

On integrating the above equation then, we get

= 1/3 x (1/a3) tan-1(t/a3) + c                        

Since, ∫1/ x+ a2 dx = 1/a tan-1(x/a) + c

Hence, I = 1/3a3 tan-1(x3/a3) + c    

Question 11. Evaluate ∫ 1/ x(x+ 1) dx

Solution:

Let us assume I = ∫ 1/ x(x+ 1) dx

= ∫ x5/ x6(x+ 1) dx                     …..(i)

Now, put x6 = t

6x5 dx = dt

x5 dx = dt/6

Now, put the above value in eq(i)

= 1/6 ∫dt/ t(t + 1)

= 1/6 ∫dt/ t+ t

= 1/6 ∫dt/ t+ 2t(1/2) + (1/2)– (1/2)2             

= 1/6 ∫dt/ (t + 1/2)– (1/2)2                 …..(ii)

Let t + 1/2 = u

dt = du 

So, put the above value in eq(ii)

= 1/6 ∫du/ (u)– (1/2)2 

On integrating the above equation then, we get

= 1/6 x 1/ 2(1/2) log|u – (1/2)/u + (1/2)| + c                    

Since ∫ 1/ x– a2dx = 1/2a log|x – a/x + a| + c

= 1/6 log|{(t + 1/2) – 1/2}/(t + 1/2) + 1/2| + c

Hence, I = 1/6 log|x6/ x+ 1| + c

Question 12. Evaluate ∫ x/ (x– x+ 1) dx

Solution:

Let us assume I = ∫ x/ (x– x+ 1) dx                  …..(i)

Let x2 = t

2x dx = dt

x dx = dt/2

Now, put the above value in eq(i)

= 1/2 ∫dt/ t– t + 1

= 1/2 ∫dt/ t– 2t(1/2) + (1/2)– (1/2)+ 1            

= 1/2 ∫dt/ (t – 1/2)+ (3/4)                   …..(ii)

Let t – 1/2 = u

dt = du

So, put the above value in eq(ii)

= 1/2 ∫du/ (u)+ (√3/2)2

On integrating the above equation then, we get

= 1/2 x 1/(√3/2) tan-1(u/(√3/2)) + c                    

Since, ∫ 1/ x+ a2dx = 1/a tan-1(x/a) + c

= 1/√3 tan-1(t – 1/2/ (√3/2)) + c

Hence, I = 1/√3 tan-1(2x– 1/ √3) + c

Question 13. Evaluate ∫ x/ (3x– 18x+ 11) dx

Solution:

Let us assume I = ∫ x/ (3x– 18x+ 11) dx

= 1/3 ∫ x/ (x– 6x+ 11/3) dx                   …..(i)

Let x2 = t

2x dx = dt

x dx = dt/2

So, put the above value in eq(i)

= 1/3 x 1/2 ∫dt/ t– 6t + 11/3

= 1/6 ∫dt/ t– 2t(3) + (3)– (3)+ 11/3            

= 1/6 ∫dt/ (t – 3)– (16/3)                      …..(ii)

Let t – 3 = u

dt = du

Now, put the above value in eq(ii)

= 1/6 ∫du/ (u)– (4/√3)2

On integrating the above equation then, we get

= 1/6 x 1/ 2(4/√3) log|u – (4/√3)/u + (4/√3)| + c                    

Since, ∫ 1/ x– a2dx = 1/2a log|x – a/x + a| + c

= √3/48 log|(t – 3 – 4/√3)/(t – 3 + 4/√3)| + c

Hence, I = √3/48 log|(x– 3 – 4/√3)/(x– 3 + 4/√3)| + c

Question 14. Evaluate ∫ ex/ (1 + ex)(2 + ex) dx

Solution:

Let us assume I = ∫ ex/ (1 + ex)(2 + ex) dx                       …..(i)

Let ex = t

ex dx = dt

So, put the above value in eq(i)

= ∫ dt/ (1 + t)(2 + t)

= ∫dt/ (1 + t) – ∫dt/(2 + t)

On integrating the above equation then, we get

= log|1 + t| – log|2 + t| + c

= log|1 + t/2 + t| + c

Hence, I = log|1 + ex/2 + ex| + c 

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