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

• Last Updated : 30 Apr, 2021

### Question 9. Evaluate âˆ« 1/ cosx+sinx dx

Solution:

Let us assume I = âˆ« 1/ cosx+sinx dx

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

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

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

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

= âˆ« 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)

= âˆ« 2dt/ 2t+1-t2

= -âˆ« 2dt/ t2-2t-1

= -âˆ« 2dt/ t2-2t+1-1-1

= -âˆ« 2dt/ (t-1)2-(âˆš2)2

= âˆ« 2dt/ (âˆš2)2-(t-1)2

Integrate the above eq. then, we get

= 2/(2âˆš2) log|âˆš2+t-1/âˆš2-t+1| +c

Hence, I = 1/âˆš2 log|âˆš2+tanx/2-1/ âˆš2-tanx/2+1| +c

### Question 10. Evaluate âˆ« 1/ 5-4cosx dx

Solution:

Let us assume I = âˆ« 1/ 5-4cosx dx

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

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

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

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

= âˆ« 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 âˆ« dt/ t2+1

On integrate the above eq. then, we get

= 2 Ã— 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 âˆ« 1/ 2+sinx+cosx dx

Solution:

Let us assume I = âˆ« 1/ 2+sinx+cosx dx

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

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

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

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

= âˆ« 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)

= âˆ« 2dt/ t2+2t+3

= âˆ« 2dt/ t2+2t+1-1+3

= 2âˆ« dt/ (t+1)2+(âˆš2)2

Integrate the above eq. then, we get

= 2/âˆš2 tan-1(t+1/ âˆš2) +c

Hence, I = âˆš2tan-1(tanx/2+1/ âˆš2) +c

### Question 12. Evaluate âˆ« 1/ sinx+âˆš3cosx dx

Solution:

Let us assume I = âˆ« 1/ sinx+âˆš3cosx dx

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

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

= âˆ« 1/ {2tan(x/2)/1+tan2x/2}+âˆš3{1-tan2(x/2)/1+tan2x/2} dx

= âˆ« 1+tan2(x/2)/ 2tan(x/2)+âˆš3-âˆš3tan2(x/2) dx

= âˆ« sec2(x/2)/ 2tan(x/2)+âˆš3-âˆš3tan2(x/2) dx (i)

Let tanx/2 = t

1/2 sec2x/2 dx = dt

Put the above value in eq. (i)

= âˆ« 2dt/ 2t+âˆš3-âˆš3t2

= -2/âˆš3 âˆ« dt/ t2-2/âˆš3t+(1/âˆš3)2-(1/âˆš3)2-1

= -2/âˆš3 âˆ« dt/ (t-1/âˆš3)2-(2/âˆš3)2

= 2/âˆš3 âˆ« dt/ (2/âˆš3)2-(t-1/âˆš3)2

Integrate the above eq. then, we get

= 2/âˆš3 x 1/2(2/âˆš3) log|2/âˆš3+t+1/âˆš3/ 2/âˆš3-t+1/âˆš3| +c

= 1/2 log|âˆš3t+1/ 3-âˆš3t|+c

Hence, I = 1/2 log|1+âˆš3+tanx/2/ 3-âˆš3tanx/2| +c

### Question 13. Evaluate âˆ« 1/ âˆš3sinx+cosx dx

Solution:

Let us assume I = âˆ« 1/ âˆš3sinx+cosx dx

âˆš3 = rcosÎ¸, and 1=rsinÎ¸

tanÎ¸=1/âˆš3

Î¸ = Ï€/6

r = âˆš3+1=2

I = âˆ« 1/ rcosÎ¸sinx+rsinÎ¸cosx dx

= 1/r âˆ« 1/ sin(x+Î¸) dx

= 1/2 âˆ« cosec(x+Î¸) dx

Integrate the above eq. then, we get

= 1/2 log|tan(x/2+Î¸/2)| +c

Hence, I = 1/2 log|tan(x/2+Ï€/12)| +c

### Question 14. Evaluate âˆ« 1/ sinx-âˆš3cosx dx

Solution:

Let us assume I = âˆ« 1/ sinx-âˆš3cosx dx

1 = rcosÎ¸, and âˆš3=rsinÎ¸

tanÎ¸=âˆš3

Î¸ = Ï€/3

r = âˆš3+1=2

I = âˆ« 1/ rcosÎ¸sinx-rsinÎ¸cosx dx

= 1/r âˆ« 1/ sin(x-Î¸) dx

= 1/2 âˆ« cosec(x-Î¸) dx

Integrate the above eq. then, we get

= 1/2 log|tan(x/2-Î¸/2)| +c

Hence, I = 1/2 log|tan(x/2-Ï€/6)| +c

### Question 15. Evaluate âˆ« 1/ 5+7cosx+sinx dx

Solution:

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

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

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

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

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

= âˆ« 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)

= âˆ« 2dt/ -2t2+12+2t

= -âˆ« dt/ t2-t-6

= – âˆ« dt/ t2-2t(1/2)+(1/2)2-(1/2)2-6

= – âˆ« 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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