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• RD Sharma Class 12 Solutions for Maths

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

### 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
=∫ex cotx 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

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