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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.13 | Set 1

• Last Updated : 19 Apr, 2021

### Question 1. Evaluate âˆ« x/ âˆšx4+a4 dx

Solution:

Let us assume I = âˆ« x/ âˆšx4+a4 dx

= âˆ« x/ âˆš(x2)2+(a2)2 dx (i)

Put x2 = t

2x dx = dt

x dx = dt/2

Put the above value in eq. (i)

= 1/2 âˆ« dt/âˆšt2 +(a2)2

Integrate the above eq. then, we get

= 1/2 log |t+ âˆšt2+(a2)2| + c [since âˆ« 1/âˆšx2+a2 dx =log|x +âˆšx2+a2| + c]

= 1/2 log |x2+ âˆš(x2)2+(a2)2| + c

Hence, I = 1/2 log |x2+ âˆšx4+a4| + c

### Question 2. Evaluate âˆ« sec2x/ âˆštan2x+4 dx

Solution:

Let us assume I =âˆ« sec2x/ âˆštan2x+4 dx (i)

Put tanx = t

sec2x dx = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆšt2+(2)2

Integrate the above eq. then, we get

= log|t +âˆšt2+(2)2| + c [since âˆ« 1/âˆšx2+a2 dx =log|x +âˆšx2+a2| + c]

= log|tanx +âˆštan2x+(2)2| + c

Hence, I = log|tanx +âˆštan2x+4| + c

### Question 3. Evaluate âˆ« ex/ âˆš16-e2x dx

Solution:

Let us assume I =âˆ« ex/ âˆš16-e2x dx (i)

Put ex = t

ex dx = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆš(4)2-(t)2

Integrate the above eq. then, we get

= sin-1(t/4) + c [ since âˆ«1/ âˆša2 – x2 dx = sin-1(x/a) + c]

= sin-1(ex/4) + c

Hence, I = sin-1(ex/4) + c

### Question 4. Evaluate âˆ« cosx/ âˆš4+sin2x dx

Solution:

Let us assume I =âˆ« cosx/ âˆš4+sin2x dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆš(2)2+t2

Integrate the above eq. then, we get

= log|t +âˆš(2)2+t2| + c [since âˆ« 1/âˆšx2+a2 dx =log|x +âˆšx2+a2| + c]

= log|sinx +âˆš(2)2+sin2x| + c

Hence, I = log|sinx +âˆš4+sin2x| + c

### Question 5. Evaluate âˆ« sinx/ âˆš4cos2x-1 dx

Solution:

Let us assume I =âˆ« sinx/ âˆš4cos2x-1 dx (i)

Put 2cosx = t

-2sinx dx = dt

sinx dx = -dt/2

Put the above value in eq. (i)

= -1/2 âˆ« dt/ âˆšt2-(1)2

Integrate the above eq. then, we get

= -1/2 log|t +âˆšt2-(1)2| + c [since âˆ« 1/âˆšx2-a2 dx =log|x +âˆšx2-a2| + c]

= -1/2 log|2cosx +âˆš(2cosx)2-(1)2| + c

Hence, I = -1/2 log|2cosx +âˆš4cos2x-1| + c

### Question 6. Evaluate âˆ« x/ âˆš4-x4 dx

Solution:

Let us assume I =âˆ« x/ âˆš4-x4 dx (i)

Put x2 = t

2x dx = dt

x dx = dt/2

Put the above value in eq. (i)

=1/2 âˆ« dt/ âˆš(2)2-(t)2

Integrate the above eq. then, we get

= sin-1(t/2) + c [ since âˆ«1/ âˆša2 – x2 dx = sin-1(x/a) + c]

= sin-1(x2/2) + c

Hence, I = sin-1(x2/2) + c

### Question 7. Evaluate âˆ« 1/ xâˆš4-9(logx)2 dx

Solution:

Let us assume I =âˆ« 1/ xâˆš4-9(logx)2 dx

=âˆ« 1/ xâˆš4-(3logx)2 dx (i)

Put 3logx = t

3/x dx = dt

1/x dx = dt/3

Put the above value in eq. (i)

=1/3 âˆ« dt/ âˆš4-t2

=1/3 âˆ« dt/ âˆš(2)2-t2

Integrate the above eq. then, we get

=1/3 sin-1(t/2) + c [since âˆ«1/ âˆša2 – x2 dx = sin-1(x/a) + c]

=1/3 sin-1(3logx/2) + c

Hence, I =1/3 sin-1(3logx/2) + c

### Question 8. Evaluate âˆ« sin8x/ âˆš9+sin44x dx

Solution:

Let us assume I =âˆ« sin8x/ âˆš9+sin44x dx (i)

Put sin24x = t

2sin4xcos4x (4)dx = dt

4sin8x dx = dt

sin8x dx = dt/4

Put the above value in eq. (i)

= 1/4 âˆ« dt/ âˆš9+t2

= 1/4 âˆ« dt/ âˆš(3)2+t2

Integrate the above eq. then, we get

= 1/4 log|t +âˆš(3)2+t2| + c [since âˆ« 1/âˆša2+x2 dx =log|x +âˆša2+x2| + c]

= 1/4 log|sin24x +âˆš(3)2+sin44x| + c

Hence, I = 1/4 log|sin24x +âˆš9+sin44x| + c

### Question 9. Evaluate âˆ« cos2x/ âˆšsin22x+8 dx

Solution:

Let us assume I =âˆ« cos2x/ âˆšsin22x+8 dx (i)

Put sin2x = t

2cos2x dx = dt

cos2x dx = dt/2

Put the above value in eq. (i)

=1/2 âˆ« dt/ âˆšt2+8

=1/2 âˆ« dt/ âˆšt2+(2âˆš2)2

Integrate the above eq. then, we get

= 1/2 log|t +âˆšt2+(2âˆš2)2| + c [since âˆ« 1/âˆšx2+a2 dx =log|x +âˆšx2+a2| + c]

= 1/2 log|sin2x +âˆšsin22x+(2âˆš2)2| + c

Hence, I = 1/2 log|sin2x +âˆšsin22x+8| + c

### Question 10. Evaluate âˆ« sin2x/ âˆšsin4x+4sin2x-2 dx

Solution:

Let us assume I =âˆ« sin2x/ âˆšsin4x+4sin2x-2 dx (i)

Put sin2x = t

2sinxcosx dx = dt

sin2x dx = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆšt2+4t-2

= âˆ« dt/ âˆšt2+2t(2)+(2)2-(2)2-2

= âˆ« dt/ âˆš(t+2)2-6 (ii)

Put t+2 =u

dt = du

Put the above value in eq. (ii)

= âˆ« du/ âˆšu2-6

= âˆ« du/ âˆšu2-(âˆš6)2

Integrate the above eq. then, we get

= log|u +âˆšu2-(âˆš6)2| + c [since âˆ« 1/âˆšx2-a2 dx =log|x +âˆšx2-a2| + c]

= log|t+2 +âˆš(t+2)2-6| + c

= log|sin2x+2 +âˆš(sin2x+2)2-6| + c

= log|sin2x+2 +âˆšsin4x+4sin2x+4-6| + c

Hence, I = log|sin2x+2 +âˆšsin4x+4sin2x-2| + c

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