# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.23 | Set 2

• Last Updated : 30 Apr, 2021

### 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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