# Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.26 | Set 1

### Question 1. âˆ«(ex(cosx -sinx))dx

Solution:

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

### Question 2. âˆ«ex(x-2+2x-3)dx

Solution:

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

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

Solution:

Given expression is
âˆ«(ex(1+sinx)/(1+cosx))dx
=âˆ«((ex(sin2(x/2)+cos2(x/2)+2sin(x/2)cos(x/2)))/(2cos2(x/2)))dx
=âˆ«(((ex(sin(x/2)+cos(x/2))2/(2cos2(x/2)))dx
=âˆ«((ex/2)(tan(x/2)+1)2)dx
=âˆ«((ex/2) (1+tan2(x/2)+2tan(x/2)))dx
=âˆ«((ex/2)(sec2(x/2)+2tan(x/2)))dx
=âˆ«((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,
âˆ«(ex)(y+(dy/dx))dx=ex(y)+c
Therefore,
âˆ«(ex((1/2)sec2(x/2)+tan(x/2)))dx
=extan(x/2) +C

### Question 4. âˆ«ex(cotx-cosec2x)dx

Solution:

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

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

Solution:

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

### Question 6. âˆ«exsecx(1+tanx)dx

Solution:

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

### Question 7. âˆ«ex(tanx -logcosx)dx

Solution:

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

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

Solution:

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

### Question 9. âˆ«ex(cotx+log sinx)dx

Solution:

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

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

Solution:

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

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