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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.23 | Set 1

Last Updated : 30 Apr, 2021
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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


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