RD Sharma Class 12 Ex 19.23 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.23 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 19.23 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

Question 1. Evaluate ∫ 1/ 5+4cosx dx

Solution:

Let us assume I = ∫ 1/ 5+4cosx dx

Put cosx = 1-tan2(x/2)/ 1+tan2(x/2)

= ∫1/ 5+4{1-tan2(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 5(1+tan2x/2)+4(1-tan2x/2) dx

= ∫ sec2(x/2)/ 5+5tan2x/2+4-4tan2x/2 dx

= ∫ sec2(x/2)/ 9+tan2x/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ (3)2+t2

On integrate the above eq. then, we get

= 2 × 1/3tan-1t/3 +c

= 2/3 tan-1{tan(x/2)/3} +c

Hence, I = 2/3 tan-1{tan(x/2)/3} +c

Question 2. Evaluate ∫ 1/ 5-4sinx dx

Solution:

Let us assume I = ∫ 1/ 5-4sinx dx

Put sinx = 2tan(x/2)/ 1+tan2(x/2)

= ∫1/ 5-4{2tan(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 5(1+tan2x/2)-4(2tanx/2) dx

= ∫ sec2(x/2)/ 5+5tan2x/2-8tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ 5t2-8t+5

= 2/5 ∫ dt/ t2-(8/5)t+1

= 2/5 ∫ dt/ t2-2t(4/5)+(4/5)2-(4/5)2+1

= 2/5 ∫ dt/ (t-4/5)2+(3/5)2

On integrate the above eq. then, we get

= 2/5 × 1/(3/5)tan-1{t-(4/5)/ (3/5)} +c

= 2/3 tan-1(5t-4)/ 3 +c

Hence, I = 2/3 tan-1(5tanx/2-4)/ 3 +c

Question 3. Evaluate ∫ 1/ 1-2sinx dx

Solution:

Let us assume I = ∫ 1/ 1-2sinx dx

Put sinx = 2tan(x/2)/ 1+tan2(x/2)

= ∫1/ 1-2{2tan(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 1(1+tan2x/2)-2(2tanx/2) dx

= ∫ sec2(x/2)/ tan2x/2-4tanx/2+1 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ t2-4t+1

= ∫ 2dt/ t2-2t(2)+(2)2-(2)2+1

= 2∫ dt/ (t-2)2+3

= 2∫ dt/ (t-2)2+(√3)2

On integrate the above eq. then, we get

= 2 × 1/2√3log|t-2-√3/ t-2+√3| +c

= 2 × 1/2√3log|tanx/2-2-√3/ tanx/2-2+√3| +c

Hence, I = 2 × 1/2√3log|tanx/2-2-√3/ tanx/2-2+√3| +c

Question 4. Evaluate ∫ 1/ 4cosx-1 dx

Solution:

Let us assume I = ∫ 1/ 4cosx-1 dx

Put cosx = 1-tan2(x/2)/ 1+tan2(x/2)

= ∫1/ 4{1-tan2(x/2)/ 1+tan2(x/2)}-1 dx

= ∫ 1+tan2(x/2)/ 4(1-tan2x/2)-(1+tan2x/2) dx

= ∫ sec2(x/2)/ 4-4tan2x/2+1-tan2x/2 dx

= ∫ sec2(x/2)/ 3-5tan2x/2 dx (i)

Let √5tanx/2 = t

√5/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ dt/ (√3)2+t2

On integrate the above eq. then, we get

= 1/√15 log|√3+t/√3-t|

Hence, I = 1/√15 log|√3+√5tan(x/2)/√3-√5tan(x/2)|

Question 5. Evaluate ∫ 1/ 1-sinx+cosx dx

Solution:

Let us assume I = ∫ 1/ 1-sinx+cosx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 1-{2tan(x/2)/1+tan2x/2} + {1-tan2(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 1+tan(x/2)-2tan(x/2)+1-tan2(x/2) dx

= ∫ sec2(x/2)/ 2-2tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/2 ∫ dt/ 1-t

= ∫ dt/ 1-t

Integrate the above eq. then, we get

= -log|1-t| +c

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

Question 6. Evaluate ∫ 1/ 3+2sinx+cosx dx

Solution:

Let us assume I = ∫ 1/ 3+2sinx+cosx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 3+2{2tan(x/2)/1+tan2x/2} + {1-tan2(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 3+3tan2(x/2)+4tan(x/2)+1-tan2(x/2) dx

= ∫ sec2(x/2)/ 2tan2x/2+4tanx/2+4 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/2 ∫ dt/ t2+2t+2

= ∫ dt/ t2+2t+1-1+2

= ∫ dt/ (t+1)2+12

Integrate the above eq. then, we get

= tan-1(t+1) +c

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

Question 7. Evaluate ∫ 1/ 13+3cosx+4sinx dx

Solution:

Let us assume I = ∫ 1/ 13+3cosx+4sinx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 13+3{1-tan2(x/2)/1+tan2(x/2)} + 4{2tan(x/2)/ 1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 13(1+tan2x/2)+3-3tan2(x/2)+8tan(x/2) dx

= ∫ sec2(x/2)/ 16+13tan2x/2-3tan2x/2+8tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/10 ∫ dt/ 16+10t2+8t

= 1/5 ∫ dt/ t2+(4/5)t+8/5

= 1/5 ∫ dt/ t+2t(2/5)2+(2/5)2-(2/5)2+8/5

= 1/5 ∫ dt/ (t+2/5)2+(6/5)2

Integrate the above eq. then, we get

= 1/5 × 1/(6/5)tan-1(t+(2/5)/ (6/5)) +c

= 1/6 tan-1(5t+2/ 6) +c

Hence, I = 1/6 tan-1(5tanx/2+2/ 6) +c

Question 8. Evaluate ∫ 1/ cosx-sinx dx

Solution:

Let us assume I = ∫ 1/ cosx-sinx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ {1-tan2(x/2)/1+tan2x/2} – {2tan(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 1-tan2(x/2)-2tan(x/2) dx

= ∫ sec2(x/2)/ 1-tan2(x/2)-2tan(x/2) dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ 1-t2-2t

= -∫ 2dt/ t2+2t-1

= -∫ 2dt/ t2+2t+1-1-1

= -∫ 2dt/ (t+1)2-(√2)2

= ∫ 2dt/ (√2)2-(t+1)2

Integrate the above eq. then, we get

= 2/(2√2) log|√2+t+1/√2-t-1| +c

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

Question 9. Evaluate ∫ 1/ cosx+sinx dx

Solution:

Let us assume I = ∫ 1/ cosx+sinx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ {1-tan2(x/2)/1+tan2x/2} + {2tan(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 2tan(x/2)+1-tan2(x/2) dx

= ∫ sec2(x/2)/ 2tan(x/2)+1-tan2(x/2) dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ 2t+1-t2

= -∫ 2dt/ t2-2t-1

= -∫ 2dt/ t2-2t+1-1-1

= -∫ 2dt/ (t-1)2-(√2)2

= ∫ 2dt/ (√2)2-(t-1)2

Integrate the above eq. then, we get

= 2/(2√2) log|√2+t-1/√2-t+1| +c

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

Question 10. Evaluate ∫ 1/ 5-4cosx dx

Solution:

Let us assume I = ∫ 1/ 5-4cosx dx

Put cosx = 1-tan2(x/2)/ 1+tan2(x/2)

= ∫1/ 5-4{1-tan2(x/2)/ 1+tan2(x/2)} dx

= ∫ 1+tan2(x/2)/ 5(1+tan2x/2)-4(1-tan2x/2) dx

= ∫ sec2(x/2)/ 5+5tan2x/2-4+4tan2x/2 dx

= ∫ sec2(x/2)/ 9tan2x/2+1 dx (i)

Let 3tanx/2 = t

3/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= 2/3 ∫ dt/ t2+1

On integrate the above eq. then, we get

= 2 × 1/3tan-1t +c

= 2/3 tan-1{3tan(x/2)} +c

Hence, I = 2/3 tan-1{3tan(x/2)} +c

Question 11. Evaluate ∫ 1/ 2+sinx+cosx dx

Solution:

Let us assume I = ∫ 1/ 2+sinx+cosx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 2+{2tan(x/2)/1+tan2x/2} + {1-tan2(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 2+2tan2(x/2)+2tan(x/2)+1-tan2(x/2) dx

= ∫ sec2(x/2)/ tan2x/2+2tanx/2+3 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ t2+2t+3

= ∫ 2dt/ t2+2t+1-1+3

= 2∫ dt/ (t+1)2+(√2)2

Integrate the above eq. then, we get

= 2/√2 tan-1(t+1/ √2) +c

Hence, I = √2tan-1(tanx/2+1/ √2) +c

Question 12. Evaluate ∫ 1/ sinx+√3cosx dx

Solution:

Let us assume I = ∫ 1/ sinx+√3cosx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ {2tan(x/2)/1+tan2x/2}+√3{1-tan2(x/2)/1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 2tan(x/2)+√3-√3tan2(x/2) dx

= ∫ sec2(x/2)/ 2tan(x/2)+√3-√3tan2(x/2) dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ 2t+√3-√3t2

= -2/√3 ∫ dt/ t2-2/√3t+(1/√3)2-(1/√3)2-1

= -2/√3 ∫ dt/ (t-1/√3)2-(2/√3)2

= 2/√3 ∫ dt/ (2/√3)2-(t-1/√3)2

Integrate the above eq. then, we get

= 2/√3 x 1/2(2/√3) log|2/√3+t+1/√3/ 2/√3-t+1/√3| +c

= 1/2 log|√3t+1/ 3-√3t|+c

Hence, I = 1/2 log|1+√3+tanx/2/ 3-√3tanx/2| +c

Question 13. Evaluate ∫ 1/ √3sinx+cosx dx

Solution:

Let us assume I = ∫ 1/ √3sinx+cosx dx

√3 = rcosθ, and 1=rsinθ

tanθ=1/√3

θ = π/6

r = √3+1=2

I = ∫ 1/ rcosθsinx+rsinθcosx dx

= 1/r ∫ 1/ sin(x+θ) dx

= 1/2 ∫ cosec(x+θ) dx

Integrate the above eq. then, we get

= 1/2 log|tan(x/2+θ/2)| +c

Hence, I = 1/2 log|tan(x/2+π/12)| +c

Question 14. Evaluate ∫ 1/ sinx-√3cosx dx

Solution:

Let us assume I = ∫ 1/ sinx-√3cosx dx

1 = rcosθ, and √3=rsinθ

tanθ=√3

θ = π/3

r = √3+1=2

I = ∫ 1/ rcosθsinx-rsinθcosx dx

= 1/r ∫ 1/ sin(x-θ) dx

= 1/2 ∫ cosec(x-θ) dx

Integrate the above eq. then, we get

= 1/2 log|tan(x/2-θ/2)| +c

Hence, I = 1/2 log|tan(x/2-π/6)| +c

Question 15. Evaluate ∫ 1/ 5+7cosx+sinx dx

Solution:

Let us assume I = ∫ 1/ 5+7cosx+sinx dx

Put sinx = 2tan(x/2)/ 1+tan2x/2

cosx = 1-tan2(x/2)/1+tan2(x/2)

= ∫ 1/ 5+7{1-tan2(x/2)/1+tan2(x/2)} + {2tan(x/2)/ 1+tan2x/2} dx

= ∫ 1+tan2(x/2)/ 5(1+tan2x/2)+7-7tan2(x/2)+2tan(x/2) dx

= ∫ sec2(x/2)/ -2tan2x/2+12+2tanx/2 dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= ∫ 2dt/ -2t2+12+2t

= -∫ dt/ t2-t-6

= – ∫ dt/ t2-2t(1/2)+(1/2)2-(1/2)2-6

= – ∫ dt/ (t-1/2)2-(5/2)2

Integrate the above eq. then, we get

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

= -1/5 log|t-3/ t+2| +c

Hence, I = 1/5 log|tanx/2+2/ tanx/2-3| +c

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