# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.26 | Set 2

Last Updated : 28 Mar, 2021

### Question 11. âˆ«ex(sin4x-4)/(2sin22x)dx

Solution:

We have,

âˆ«ex(sin4x-4)/(2sin22x)dx

=âˆ«ex(2sin2xcos2x-4)/(2sin22x)dx

=âˆ«ex(((2sin2xcos2x)/(2sin22x))-4/(2sin22x))dx

=âˆ«ex(cot2x-2cosec22x)dx

=âˆ«excot2xdx-2âˆ«excosec22xdx

Integrating by parts,

excot2x-2âˆ«exd(cot2x)/dx-2âˆ«excosec22xdx

= excot2x+2âˆ«excosec22xdx-2âˆ«excosec22xdx

= excot2x+C

### Question 12. âˆ«ex(2-x)/(1-x)2dx

We have,

âˆ«ex(2-x)/(1-x)2dx

=âˆ«ex((1-x)+1)/(1-x)2dx

=âˆ«ex(((1/(1-x))+(1/(1-x)2)))dx

=âˆ«ex(1/(1-x))+âˆ«ex(1/(1-x)2)dx

= ex/(1-x)+C

### Question 13. âˆ«ex(1+x)/(x+2)2dx

We have,

âˆ«ex(1+x)/(x+2)2dx

=âˆ«ex((2+x)-1)/(x+2)2dx

=âˆ«ex(((2+x)/(x+2)2)-1/(x+2)2)dx

=âˆ«ex((1/(x+2))-1/(x+2)2)dx

= ex/(x+2)dx

### Question 14. âˆ«e(-x/2)(1-sinx)(1/2)/(1+cosx)dx

We have,

âˆ«e(-x/2)(1-sinx)(1/2)/(1+cosx)dx

Let x/2 = t

So, x = 2t

So the equation is,

âˆ«2e(-t)(1-sin2t)(1/2)/(1+cos2t)dt

=2âˆ«e(-t)(sin2t+cos2t-2sintcost)(1/2)/(2cos2(t))dt

=âˆ«e(-t)(sint-cost)(2*1/2)/cos2tdt

=âˆ«e(-t)(sint-cost)/cos2tdt

=âˆ«e(-t)(tantsect-sect)dt

=âˆ«e(-t)(tant sect)dt-âˆ«e(-t)sectdt

=âˆ«e(-t)(tant sect)dt -e(-t)sect -âˆ«e(-t)(d(sect)/dt) dt

=âˆ«e(-t)(tant sect)dt -e(-t)sect -âˆ«e(-t)sect tantdt

= e(-t)sect

= e(-x/2)sec(x/2)+C

### Question 15. âˆ«ex(logx +1/x)dx

Solution:

We have,

âˆ«ex(logx +1/x)dx

= ex(logx)+C

### Question 16. âˆ«ex(logx+1/x2)dx

Solution:

We have,

âˆ«ex(logx +1/x2)dx

=âˆ«ex(logx+1/x-1/x+1/x2)dx

=âˆ«ex((logx-1/x)+((1/x)+(1/x2))dx

= ex(logx-1/x)+C

### Question 17. âˆ«ex/x(x(logx)2+2logx)dx

Solution:

We have,

âˆ«ex/x(x(logx)2+2logx)dx

=âˆ«ex((logx)2+2ex(logx)/x)dx

=âˆ«ex(logx)2dx +âˆ«(2ex/x)(logx)dx

Integrating by parts,

= ex(logx)2-âˆ«ex(d(logx)2/dx)dx +âˆ«(2ex/x)(logx)dx

= ex(logx)2-âˆ«(ex/x)2logxdx+âˆ«2ex/x(logx)dx

= ex(logx)2+C

### Question 18. âˆ«ex(sin-1x+1/(1-x2)1/2)dx

Solution:

= exsin-1x-âˆ«ex(d(sin-1x)/dx) dx +âˆ«ex/(1-x2)1/2dx

= exsin-1x- âˆ«ex/(1-x2)1/2dx+âˆ«ex/(1-x2)1/2dx

= exsin-1x+C

### Question 19. âˆ«e2x(-sinx +2cosx)dx

Solution:

= -âˆ«e2xsinxdx +2âˆ«e2xcosxdx

= -âˆ«e2xsinxdx+2((1/2)e2xcosx+âˆ«(1/2)e2xsinxdx)

= -âˆ«e2xsinxdx+e2xcosx+âˆ«e2xsinxdx

= e2xcosx+C

### Question 20. âˆ«ex(tan-1x+1/(1+x2))dx

Solution:

= extan-1x-âˆ«ex (d(tan-1x)/dx) dx+âˆ«ex/(1+x2)dx

= extan-1x-âˆ«ex/(1+x2)dx+âˆ«ex/(1+x2)dx

= extan-1x+C

### Question 21. âˆ«ex((sinxcosx-1)/sin2x)dx

Solution:

= âˆ«ex(cotx-cosec2x)dx

= excotx-âˆ«ex(d(cotx)/dx) dx-âˆ«excosec2xdx

= excotx+âˆ«excosec2xdx -âˆ«excosec2xdx

= excotx+C

### Question 22. âˆ«(tan(logx)+sec2(logx))dx

Solution:

Suppose,

logx=z

=>ez=x

=>d(ez)/dx=1

=>ezdz=dx

Substituting it in original question,

âˆ«(tan z+sec2z)ezdz

= eztanz -âˆ«ez(d(tanz)/dz))dz+âˆ«ezsec2zdz

= eztanz-âˆ«ezsec2zdz+âˆ«ezsec2zdz

= eztanz+C

= e(logx)tan(logx)+C

= x tan(logx)+C

### Question 23. âˆ«ex(x-4)/(x-2)3dx

Solution:

= âˆ«ex((x-2)-2)/(x-2)3dx

= âˆ«ex((1/(x-2)2)-(2/(x-2)3))dx

= ex/(x-2)2-âˆ«ex(d((x-2)-2)/dx)dx-2âˆ«ex/(x-2)3dx

= ex/(x-2)2+2âˆ«ex/(x-2)3dx -2âˆ«ex/(x-2)3dx

= ex/(x-2)2+C

### Question 24. âˆ«e2x((1-sin2x)/(1-2cosx))dx

Solution:

=âˆ«e2x((1-sin2x)/2sin2x)dx

=âˆ«e2x((cosec2x/2)-cotx)dx

Suppose,

I=I1+I2

I1=1/2âˆ«e2xcosec2xdx

I2=-âˆ«e2xcotxdx

let,

u=e2x

=du=2e2xdx

and

âˆ«cosec2xdx=âˆ«dv

=>v=-cotx+C

So,

I1=1/2[e2x(-cotx)-âˆ«(-cotx)2e2xdx]

I1=1/2(e2x(-cotx))+âˆ«cotxe2xdx

Thus,

I=(1/2)(e2x(-cotx))+âˆ«cotxe2xdx -âˆ«e2x cotx dx

=>I=(1/2)(e2x(-cotx))+C

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