RD Sharma Class 12 Ex 19.18 Solutions Chapter 19 Indefinite Integrals

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RD Sharma Class 12 Ex 19.18 Solutions Chapter 19 Indefinite

Question 1. Evaluate ∫ x/ √x⁴+a⁴ dx

Solution:

Let us assume I = ∫ x/ √x⁴+a⁴ dx

= ∫ x/ √(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 ∫ dt/√t² +(a²)

Integrate the above eq. then, we get

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

= 1/2 log |x²+ √(x²)²+(a²)²| + c

Hence, I = 1/2 log |x²+ √x⁴+a⁴| + c

Question 2. Evaluate ∫ sec²x/ √tan²x+4 dx

Solution:

Let us assume I =∫ sec²x/ √tan²x+4 dx (i)

Put tan x = t

sec²x dx = dt

Put the above value in eq. (i)

= ∫ dt/ √t²+(2)²

Integrate the above eq. then, we get

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

= log|tanx +√tan²x+(2)²| + c

Hence, I = log|tanx +√tan²x+4| + c

Question 3. Evaluate ∫ e^x/ √16-e^2x dx

Solution:

Let us assume I =∫ e^x/ √16-e^2x dx (i)

Put e^x = t

e^x dx = dt

Put the above value in eq. (i)

= ∫ dt/ √(4)²-(e)²

Integrate the above eq. then, we get

= sin^-1(t/4) + c [since ∫1/ √a² – x² dx = sin^-1(x/a) + c]

= sin^-1(e^x/4) + c

Hence, I = sin^-1(e^x/4) + c

Question 4. 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+t²

Integrate the above eq. then, we get

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

= log|sinx +√(2)²+sin²x| + c

Hence, I = log|sinx +√4+sin²x| + c

Question 5. Evaluate ∫ sinx/ √4cos²x-1 dx

Solution:

Let us assume I =∫ sinx/ √4cos²x-1 dx (i)

Put 2cosx = t

-2sinx dx = dt

sinx dx = -dt/2

Put the above value in eq. (i)

= -1/2 ∫ dt/ √t²-(1)²

Integrate the above eq. then, we get

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

= -1/2 log|2cosx +√(2cosx)²-(1)²| + c

Hence, I = -1/2 log|2cosx +√4cos²x-1| + c

Question 6. Evaluate ∫ x/ √4-x⁴ dx

Solution:

Let us assume I =∫ x/ √4-x⁴ dx (i)

Put x² = t

2x dx = dt

x dx = dt/2

Put the above value in eq. (i)

=1/2 ∫ 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(x²/2) + c

Hence, I = sin^-1(x²/2) + c

Question 7. Evaluate ∫ 1/ x√4-9(logx)² dx

Solution:

Let us assume I =∫ 1/ x√4-9(logx)² dx

=∫ 1/ x√4-(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 ∫ dt/ √4-t²

=1/3 ∫ dt/ √(2)²-t²

Integrate the above eq. then, we get

=1/3 sin^-1(t/2) + c [since ∫1/ √a² – 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 ∫ sin8x/ √9+sin⁴4x dx

Solution:

Let us assume I =∫ sin8x/ √9+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 ∫ dt/ √9+t²

= 1/4 ∫ dt/ √(3)²+t²

Integrate the above eq. then, we get

= 1/4 log|t +√(3)²+t²| + c [since ∫ 1/√a²+x² dx =log|x +√a²+x²| + c]

= 1/4 log|sin⁴4x +√(3)²+sin⁴4x| + c

Hence, I = 1/4 log|sin²4x +√9+sin⁴4x| + c

Question 9. Evaluate ∫ cos2x/ √sin²2x+8 dx

Solution:

Let us assume I =∫ cos2x/ √sin²2x+8 dx (i)

Put sin2x = t

2cos2x dx = dt

cos2x dx = dt/2

Put the above value in eq. (i)

=1/2 ∫ dt/ √t²+8

=1/2 ∫ dt/ √t²+(2√2)²

Integrate the above eq. then, we get

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

= 1/2 log|sin2x +√sin²2x+(2√2)²| + c

Hence, I = 1/2 log|sin2x +√sin²2x+8| + c

Question 10. Evaluate ∫ sin2x/ √sin⁴x+4sin²x-2 dx

Solution:

Let us assume I =∫ sin2x/ √sin⁴x+4sin²x-2 dx (i)

Put sin²x = t

2sinxcosx dx = dt

sin2x dx = dt

Put the above value in eq. (i)

= ∫ dt/ √t²+4t-2

= ∫ dt/ √t²+2t(2)+(2)²-(2)²-2

= ∫ dt/ √(t+2)²-6 (ii)

Put t+2 =u

dt = du

Put the above value in eq. (ii)

= ∫ du/ √u²-6

= ∫ du/ √u²-(√6)²

Integrate the above eq. then, we get

= log|u +√u²-(√6)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= log|t+2 +√(t+2)²-6| + c

= log|sin²x+2 +√(sin²x+2)²-6| + c

= log|sin²x+2 +√sin⁴x+4sin²x+4-6| + c

Hence, I = log|sin²x+2 +√sin⁴x+4sin²x-2| + c

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/ x^2/3√x^2/3-4 dx

Solution:

Let us assume I =∫ 1/ x^2/3√x^2/3-4 dx (i)

Put x^1/3 = t

(1/3) x^1/3-1 dx = dt

(1/3) x^-2/3 dx = dt

dx/ x^2/3 = 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|x^1/3 +√(x^1/3)²-(2)²| + c

Hence, I = 3 log|x^1/3 +√x^2/3– 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

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