# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.13 | Set 2

Last Updated : 19 Apr, 2021

### Question 11. Evaluate âˆ« sin2x/ âˆšcos4x-sin2x+2 dx

Solution:

Let us assume I =âˆ« sin2x/ âˆšcos4x-sin2x+2 dx

=âˆ« sin2x/ âˆšcos4x-(1-cos2x)+2 dx (i)

Put cos2x = t

-2sinxcosx dx = dt

sin2x dx = -dt

Put the above value in eq. (i)

= -âˆ« dt/ âˆšt2-(1-t)+2

= -âˆ« dt/ âˆšt2-1+t+2

= -âˆ« dt/ âˆšt2+t+1

= -âˆ« dt/ âˆšt2+t+(1/4)+(3/4)

= -âˆ« dt/ âˆš(t+1/2)2+ 3/4 (ii)

Put t+1/2 =u

dt = du

Put the above value in eq. (ii)

= -âˆ« du/ âˆšu2+ 3/4

= -âˆ« du/ âˆšu2+3/4

Integrate the above eq. then, we get

= -log|u +âˆšu2+3/4| + c [since âˆ« 1/âˆšx2+a2 dx =log|x +âˆšx2+a2| + c]

= -log|t+1/2 +âˆš(t+1/2)2+3/4| + c

= -log|t+1/2 +âˆš(t2+t+1)| + c

= -log|(cos2x+1/2) +âˆš(cos4x+cos2x+1| + c

Hence, I = -log|(cos2x+1/2) +âˆš(cos4x+cos2x+1| + c

### Question 12. Evaluate âˆ« cosx/ âˆš4-sin2x dx

Solution:

Let us assume I =âˆ« cosx/ âˆš4-sin2x dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆš(2)2-(t)2

Integrate the above eq. then, we get

= sin-1(t/2) + c [ since âˆ«1/ âˆša2 – x2 dx = sin-1(x/a) + c]

= sin-1(sinx/2) + c

Hence, I = sin-1(sinx/2) + c

### Question 13. Evaluate âˆ« 1/ x2/3âˆšx2/3-4 dx

Solution:

Let us assume I =âˆ« 1/ x2/3âˆšx2/3-4 dx (i)

Put x1/3 = t

(1/3) x1/3-1 dx = dt

(1/3) x-2/3 dx = dt

dx/ x2/3 = 3dt

Put the above value in eq. (i)

= 3 âˆ« dt/ âˆšt2-(2)2

Integrate the above eq. then, we get

= 3 log|t +âˆšt2-(2)2| + c [since âˆ« 1/âˆšx2-a2 dx =log|x +âˆšx2-a2| + c]

= 3 log|x1/3 +âˆš(x1/3)2-(2)2| + c

Hence, I = 3 log|x1/3 +âˆšx2/3-4| + c

### Question 14. Evaluate âˆ« 1/ âˆš(1-x2)[9+(sin-1x)2 dx

Solution:

Let us assume I =âˆ« 1/ âˆš(1-x2)[9+(sin-1x)2 dx (i)

Put sin-1x = t

dx/âˆš1-x2 = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆš(3)2+t

Integrate the above eq. then, we get

= log|t +âˆš(3)2+t2| + c [since âˆ« 1/âˆša2+x2 dx =log|x +âˆša2+x2| + c]

= log|sin-1x +âˆš(3)2+(sin-1x)2| + c

Hence, I = log|sin-1x +âˆš9+(sin-1x)2| + c

### Question 15. Evaluate âˆ« cosx/ âˆšsin2x-2sinx-3 dx

Solution:

Let us assume I =âˆ« cosx/ âˆšsin2x-2sinx-3 dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= âˆ« dt/ âˆšt2-2t-3

= âˆ« dt/ âˆšt2-2t+(1)2-(1)2-3

= âˆ« dt/ âˆš(t-1)2-(2)2 (ii)

Put t-1 =u

dt = du

Put the above value in eq. (ii)

= âˆ« du/ âˆšu2-(2)2

Integrate the above eq. then, we get

= log|u +âˆšu2-(2)2| + c [since âˆ« 1/âˆšx2-a2 dx =log|x +âˆšx2-a2| + c]

= log|t-1 +âˆš(t-1)2-4| + c

= log|t-1 +âˆšt2-2t+1-4| + c

= log|t-1 +âˆšt2-2t-3| + c

= log|sinx-1 +âˆšsin2x-2sinx-3| + c

Hence, I = log|sinx-1 +âˆšsin2x-2sinx-3| + c

### Question 16. Evaluate âˆ« âˆšcosecx-1 dx

Solution:

Let us assume I =âˆ« âˆšcosecx-1 dx

= âˆ« âˆš1/sinx -1 dx

=âˆ« âˆš1-sinx /sinx dx

=âˆ« âˆš(1-sinx)+(1+sinx) /(1+sinx)sinx dx

=âˆ« âˆš(1+sinx-sinx-sin2x) /sin2x+sinx dx

=âˆ« âˆšcos2x /sin2x+sinx dx

=âˆ« cosx /âˆšsin2x+sinx dx (i)

Let sinx = t

cosx dx = dt

Put the above value in eq. (i)

= âˆ« dt/âˆšt2+t

= âˆ« dt/âˆšt2+2t(1/2)+(1/2)2-(1/2)2

= âˆ« dt/âˆš(t+1/2)2-(1/2)2 (ii)

Let t+1/2 = u

dt = du

Put the above value in eq. (ii)

= âˆ« dt/âˆš(u)2-(1/2)2

Integrate the above eq. then, we get

= log|u +âˆšu2-(1/2)2| + c [since âˆ« 1/âˆšx2-a2 dx =log|x +âˆšx2-a2| + c]

= log|t+1/2 +âˆš(t+1/2)2-(1/2)2| + c

= log|t+1/2 +âˆšt2+t| + c

Hence, I = log|sinx+1/2 +âˆšsin2x+sinx| + c

### Question 17. Evaluate âˆ« sinx-cosx/ âˆšsin2x dx

Solution:

Let us assume I =âˆ« sinx-cosx/ âˆšsin2x dx

âˆ« sinx-cosx/ âˆšsin2x dx = âˆ« sinx-cosx/ âˆš(sinx+cosx)2 -1 dx

= âˆ« sinx-cosx/ âˆš(sinx+cosx)2 -1 dx (i)

Let sinx+cosx = t

cosx-sinx dx = dt

Put the above value in eq. (i)

= -âˆ« dt/ âˆšt2-(1)2

Integrate the above eq. then, we get

= – log|t +âˆšt2-(1)2| + c [since âˆ« 1/âˆšx2-a2 dx =log|x +âˆšx2-a2| + c]

= – log|sinx+cosx +âˆšsin2x| + c

Hence, I = – log|sinx+cosx +âˆšsin2x| + c

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