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• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.18 | Set 2

### Question 11. Evaluate ∫ sin2x/ √cos4x-sin2x+2 dx

Solution:

Let us assume I =∫ sin2x/ √cos4x-sin2x+2 dx

=∫sin2x/ √cos4x-(1-cos2x)+2 dx (i)

Put cos2x = t

-2sinxcosx dx = dt

sin2x dx = -dt

Put the above value in eq. (i)

= -∫ dt/ √t2-(1-t)+2

= -∫ dt/ √t2-1+t+2

= -∫ dt/ √t2+t+1

= -∫ dt/ √t2+t+(1/4)+(3/4)

= -∫ dt/ √(t+1/2)2+ 3/4 (ii)

Put t+1/2 =u

dt = du

Put the above value in eq. (ii)

= -∫ du/ √u2+ 3/4

= -∫ du/ √u2+3/4

Integrate the above eq. then, we get

= -log|u +√u2+3/4| + c [since ∫ 1/√x2+a2 dx =log|x +√x2+a2| + c]

= -log|t+1/2 +√(t+1/2)2+3/4| + c

= -log|t+1/2 +√(t2+t+1)| + c

= -log|(cos2x+1/2) +√(cos4x+cos2x+1| + c

Hence, I = -log|(cos2x+1/2) +√(cos4x+cos2x+1| + c

### Question 12. Evaluate ∫ cosx/ √4-sin2x dx

Solution:

Let us assume I =∫ cosx/ √4-sin2x dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/ √(2)2-(t)2

Integrate the above eq. then, we get

= sin-1(t/2) + c [since ∫1/ √a2 – x2 dx = sin-1(x/a) + c]

= sin-1(sinx/2) + c

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

### Question 13. Evaluate ∫ 1/ x2/3√x2/3-4 dx

Solution:

Let us assume I =∫ 1/ x2/3√x2/3-4 dx (i)

Put x1/3 = t

(1/3) x1/3-1 dx = dt

(1/3) x-2/3 dx = dt

dx/ x2/3 = 3dt

Put the above value in eq. (i)

= 3 ∫ dt/ √t2-(2)2

Integrate the above eq. then, we get

= 3 log|t +√t2-(2)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]

= 3 log|x1/3 +√(x1/3)2-(2)2| + c

Hence, I = 3 log|x1/3 +√x2/3– 4| + c

### Question 14. Evaluate ∫ 1/ √(1-x2)[9+(sin-1x)2 dx

Solution:

Let us assume I =∫ 1/ √(1-x2)[9+(sin-1x)2 dx (i)

Put sin-1x = t

dx/√1-x2 = dt

Put the above value in eq. (i)

= ∫ dt/ √(3)2 + t

Integrate the above eq. then, we get

= log|t +√(3)2+t2| + c [since ∫ 1/√a2+x2 dx =log|x +√a2+x2| + c]

= log|sin-1x +√(3)2+(sin-1x)2| + c

Hence, I = log|sin-1x +√9+(sin-1x)2| + c

### Question 15. Evaluate ∫ cosx/ √sin2x-2sinx-3 dx

Solution:

Let us assume I =∫ cosx/ √sin2x-2sinx-3 dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/ √t2-2t-3

= ∫ dt/ √t2-2t+(1)2-(1)2-3

= ∫ dt/ √(t-1)2-(2)3 (ii)

Put t-1 =u

dt = du

Put the above value in eq. (ii)

= ∫ du/ √u2-(2)2

Integrate the above eq. then, we get

= log|u +√u2-(2)2| + c [since ∫1/√x2-a2 dx =log|x +√x2-a2| + c]

= log|t-1 +√(t-1)2-4| + c

= log|t-1 +√t2-2t+1-4| + c

= log|t-1 +√t2-2t-3| + c

= log|sinx-1 +√sin2x-2sinx-3| + c

Hence, I = log|sinx-1 +√sin2x-2sinx-3| + c

### Question 16. Evaluate ∫ √cosecx-1 dx

Solution:

Let us assume I =∫ √cosecx-1 dx

= ∫ √1/sinx -1 dx

=∫ √1-sinx /sinx dx

=∫ √(1-sinx)+(1 + sinx) /(1+sinx)sinx dx

=∫ √(1+sinx-sinx-sin2x) /sin2x+sinx dx

=∫ √cos2x /sin2x+sinx dx

=∫ cosx /√sin2x+sinx dx (i)

Let sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/√t2+t

= ∫ dt/√t2+2t(1/2)+(1/2)2-(1/2)2

= ∫ dt/√(t+1/2)2-(1/2)2 (ii)

Let t+1/2 = u

dt = du

Put the above value in eq. (ii)

= ∫ dt/√(u)2-(1/2)2

Integrate the above eq. then, we get

= log|u +√u2-(1/2)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]

= log|t+1/2 +√(t+1/2)2-(1/2)2| + c

= log|t+1/2 +√t2+t| + c

Hence, I = log|sinx+1/2 +√sin2x+sinx| + c

### Question 17. Evaluate ∫ sinx-cosx/ √sin2x dx

Solution:

Let us assume I =∫ sinx-cosx/ √sin2x dx

∫ sinx-cosx/ √sin2x dx = ∫ sinx-cosx/ √(sinx+cosx)2 -1 dx

= ∫ sinx-cosx/ √(sinx+cosx)2 -1 dx (i)

Let sinx+cosx = t

cosx-sinx dx = dt

Put the above value in eq. (i)

= -∫ dt/ √t2-(1)2

Integrate the above eq. then, we get

= – log|t +√t2-(1)2| + c [since ∫ 1/√x2-a2 dx =log|x +√x2-a2| + c]

= – log|sinx+cosx +√sin2x| + c

Hence, I = – log|sinx+cosx +√sin2x| + c