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

  • Last Updated : 28 Mar, 2021

Evaluate the following integrals.

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

Question 15. ∫ex(logx +1/x)dx

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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