# RD Sharma Class 12 Ex 19.26 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.26 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.26 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.26 Solutions Chapter 19 Indefinite Integrals

### Question 1. ∫(ex(cosx -sinx))dx

Solution:

Given expression is
∫(excosx)-(exsinx)dx
=∫(excosx) dx -∫(exsinx)dx
=ex(cosx )-∫(exd(cosx)/dx-∫exsinx dx
=ex(cosx )+∫exsinx dx-∫exsinx dx
=ex(cosx) + c

### Question 2. ∫ex(x-2+2x-3)dx

Solution:

Given expression is
∫ex(x-2+2x-3)dx
=∫exx-2dx +∫ex(2x-3)dx
=exx-2-∫ex(d(x-2)/dx)dx +2∫exx-3dx
=exx-2+2∫exx-3dx +2∫exx-3dx
=exx-2+C

### Question 3. ∫(ex(1+sinx)/(1+cosx))dx

Solution:

Given expression is
∫(ex(1+sinx)/(1+cosx))dx
=∫((ex(sin2(x/2)+cos2(x/2)+2sin(x/2)cos(x/2)))/(2cos2(x/2)))dx
=∫(((ex(sin(x/2)+cos(x/2))2/(2cos2(x/2)))dx
=∫((ex/2)(tan(x/2)+1)2)dx
=∫((ex/2) (1+tan2(x/2)+2tan(x/2)))dx
=∫((ex/2)(sec2(x/2)+2tan(x/2)))dx
=∫((ex)((1/2)sec2(x/2)+tan(x/2)))dx
Suppose tan(x/2)=y
=>dy/dx=d(tan(x/2))/dx
=>dy/dx=(1/2)(sec2(x/2))
So, the above expression becomes,
∫(ex)(y+(dy/dx))dx=ex(y)+c
Therefore,
∫(ex((1/2)sec2(x/2)+tan(x/2)))dx
=extan(x/2) +C

### Question 4. ∫ex(cotx-cosec2x)dx

Solution:

Given expression is
∫(ex(cotx – cosec2x))dx
=∫ecotx dx -∫excosec2xdx
=excotx-∫(ex(d(cot x)/dx))dx-∫excosec2xdx
=excotx+∫excosec2xdx -∫excosec2xdx
=excotx +c

### Question 5. ∫(ex((1/2x)-(1/2x2)))dx

Solution:

Given expression is,
∫(ex((1/2x)-(1/2x2)))dx
=∫ex(1/2x)dx-∫ex(1/2x2)dx
=(ex/2x)-∫ex(d(1/2x)/dx)dx -∫ex(1/2x2)dx
=(ex/2x)+∫(ex/2x2)dx-∫(ex/2x2)dx
=ex/2x+c

### Question 6. ∫exsecx(1+tanx)dx

Solution:

Given expression is,
∫exsecx(1+tanx)dx
=∫exsecxdx+∫ex(secx)(tanx)dx
=ex(secx)-∫ex(d(sec x tan x)/dx) +∫exsecx tanx dx
=ex(secx)+c

### Question 7. ∫ex(tanx -logcosx)dx

Solution:

Given expression is,
∫ex(tanx -logcosx)dx
=∫ex(tanx)dx -∫ex(logcosx)dx
=∫ex(tanx)dx- exlogcosx +∫ex(d(log cosx)/dx)dx
=∫ex(tanx)dx- exlogcosx -∫extanxdx
=-exlogcosx +c
=exlog(secx)+c

### Question 8. ∫ex[secx+log(secx +tanx)]dx

Solution:

Given expression is,
∫ex[secx+log(secx +tanx)]dx
=∫ex(secx)dx+∫exlog(secx+tanx)dx
=∫ex(secx)dx+exlog(secx+tanx)-∫ex(d(log(secx+tanx))/dx)dx
=∫ex(secx)dx+ex(log(secx+tanx))-∫exsecxdx
=ex(log(secx+tanx))+c

### Question 9. ∫ex(cotx+log sinx)dx

Solution:

Given expression is,
∫ex(cotx+log sinx)dx
=∫ex(cotx)dx+∫ex(log sinx)dx
=∫ex(cotx)dx+ex(log(sinx))-∫ex(d(log sinx)/dx)dx
=∫ex(cotx)dx + ex(log sinx) -∫excotx dx
=ex(log sinx)+c

### Question 10. ∫ex((x+1-2)/(x+1)3)dx

Solution:

Given expression is,
∫ex((x+1-2)/(x+1)3)dx
=∫ex((1/(x+1)2)-(2/(x+1)3))dx
=∫ex(1/(x+1)2)dx-∫(2ex)/(x+1)3dx
=ex/(x+1)2-∫ex(d(1/(x+1)2)/dx)-∫(2ex)/(x+1)3dx
=ex/(x+1)2-∫(ex(-2)/(x+1)3)dx -∫(2ex)/(x+1)3dx
=ex/(x+1)2+∫(ex(2)/(x+1)3)dx -∫(2ex)/(x+1)3dx
=ex/(x+1)2+c

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

Solution:

We have,

∫ex(sin4x-4)/(2sin22x)dx

=∫ex(2sin2xcos2x-4)/(2sin22x)dx

=∫ex(((2sin2xcos2x)/(2sin22x))-4/(2sin22x))dx

=∫ex(cot2x-2cosec22x)dx

=∫excot2xdx-2∫excosec22xdx

Integrating by parts,

excot2x-2∫exd(cot2x)/dx-2∫excosec22xdx

= excot2x+2∫excosec22xdx-2∫excosec22xdx

= excot2x+C

### Question 12. ∫ex(2-x)/(1-x)2dx

We have,

∫ex(2-x)/(1-x)2dx

=∫ex((1-x)+1)/(1-x)2dx

=∫ex(((1/(1-x))+(1/(1-x)2)))dx

=∫ex(1/(1-x))+∫ex(1/(1-x)2)dx

= ex/(1-x)+C

### Question 13. ∫ex(1+x)/(x+2)2dx

We have,

∫ex(1+x)/(x+2)2dx

=∫ex((2+x)-1)/(x+2)2dx

=∫ex(((2+x)/(x+2)2)-1/(x+2)2)dx

=∫ex((1/(x+2))-1/(x+2)2)dx

= ex/(x+2)dx

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

We have,

∫e(-x/2)(1-sinx)(1/2)/(1+cosx)dx

Let x/2 = t

So, x = 2t

So the equation is,

∫2e(-t)(1-sin2t)(1/2)/(1+cos2t)dt

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

=∫e(-t)(sint-cost)(2*1/2)/cos2tdt

=∫e(-t)(sint-cost)/cos2tdt

=∫e(-t)(tantsect-sect)dt

=∫e(-t)(tant sect)dt-∫e(-t)sectdt

=∫e(-t)(tant sect)dt -e(-t)sect -∫e(-t)(d(sect)/dt) dt

=∫e(-t)(tant sect)dt -e(-t)sect -∫e(-t)sect tantdt

= e(-t)sect

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

Solution:

We have,

∫ex(logx +1/x)dx

= ex(logx)+C

### Question 16. ∫ex(logx+1/x2)dx

Solution:

We have,

∫ex(logx +1/x2)dx

=∫ex(logx+1/x-1/x+1/x2)dx

=∫ex((logx-1/x)+((1/x)+(1/x2))dx

= ex(logx-1/x)+C

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

Solution:

We have,

∫ex/x(x(logx)2+2logx)dx

=∫ex((logx)2+2ex(logx)/x)dx

=∫ex(logx)2dx +∫(2ex/x)(logx)dx

Integrating by parts,

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

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

= ex(logx)2+C

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

Solution:

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

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

= exsin-1x+C

### Question 19. ∫e2x(-sinx +2cosx)dx

Solution:

= -∫e2xsinxdx +2∫e2xcosxdx

= -∫e2xsinxdx+2((1/2)e2xcosx+∫(1/2)e2xsinxdx)

= -∫e2xsinxdx+e2xcosx+∫e2xsinxdx

= e2xcosx+C

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

Solution:

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

= extan-1x-∫ex/(1+x2)dx+∫ex/(1+x2)dx

= extan-1x+C

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

Solution:

= ∫ex(cotx-cosec2x)dx

= excotx-∫ex(d(cotx)/dx) dx-∫excosec2xdx

= excotx+∫excosec2xdx -∫excosec2xdx

= excotx+C

### Question 22. ∫(tan(logx)+sec2(logx))dx

Solution:

Suppose,

logx=z

=>ez=x

=>d(ez)/dx=1

=>ezdz=dx

Substituting it in original question,

∫(tan z+sec2z)ezdz

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

= eztanz-∫ezsec2zdz+∫ezsec2zdz

= eztanz+C

= e(logx)tan(logx)+C

= x tan(logx)+C

### Question 23. ∫ex(x-4)/(x-2)3dx

Solution:

= ∫ex((x-2)-2)/(x-2)3dx

= ∫ex((1/(x-2)2)-(2/(x-2)3))dx

= ex/(x-2)2-∫ex(d((x-2)-2)/dx)dx-2∫ex/(x-2)3dx

= ex/(x-2)2+2∫ex/(x-2)3dx -2∫ex/(x-2)3dx

= ex/(x-2)2+C

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

Solution:

=∫e2x((1-sin2x)/2sin2x)dx

=∫e2x((cosec2x/2)-cotx)dx

Suppose,

I=I1+I2

I1=1/2∫e2xcosec2xdx

I2=-∫e2xcotxdx

let,

u=e2x

=du=2e2xdx

and

∫cosec2xdx=∫dv

=>v=-cotx+C

So,

I1=1/2[e2x(-cotx)-∫(-cotx)2e2xdx]

I1=1/2(e2x(-cotx))+∫cotxe2xdx

Thus,

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

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

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