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

Evaluate the following integrals.

Question 11. ∫e^x(sin4x-4)/(2sin²2x)dx

Solution: 

We have,

 ∫e^x(sin4x-4)/(2sin²2x)dx

=∫e^x(2sin2xcos2x-4)/(2sin²2x)dx

=∫e^x(((2sin2xcos2x)/(2sin²2x))-4/(2sin²2x))dx

=∫e^x(cot2x-2cosec²2x)dx

=∫e^xcot2xdx-2∫e^xcosec²2xdx

Integrating by parts,

e^xcot2x-2∫e^xd(cot2x)/dx-2∫e^xcosec²2xdx 

= e^xcot2x+2∫e^xcosec²2xdx-2∫e^xcosec²2xdx

= e^xcot2x+C

Question 12. ∫e^x(2-x)/(1-x)²dx

We have,

∫e^x(2-x)/(1-x)²dx

=∫e^x((1-x)+1)/(1-x)²dx

=∫e^x(((1/(1-x))+(1/(1-x)²)))dx

=∫e^x(1/(1-x))+∫e^x(1/(1-x)²)dx

= e^x/(1-x)+C

Question 13. ∫e^x(1+x)/(x+2)²dx

We have,

∫e^x(1+x)/(x+2)²dx

=∫e^x((2+x)-1)/(x+2)²dx

=∫e^x(((2+x)/(x+2)²)-1/(x+2)²)dx

=∫e^x((1/(x+2))-1/(x+2)²)dx

= e^x/(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. ∫e^x(logx +1/x)dx

Solution:

We have,

∫e^x(logx +1/x)dx

= e^x(logx)+C

Question 16. ∫e^x(logx+1/x²)dx

Solution:

We have,

∫e^x(logx +1/x²)dx

=∫e^x(logx+1/x-1/x+1/x²)dx

=∫e^x((logx-1/x)+((1/x)+(1/x²))dx

= e^x(logx-1/x)+C

Question 17. ∫e^x/x(x(logx)²+2logx)dx

Solution:

We have,

∫e^x/x(x(logx)²+2logx)dx

=∫e^x((logx)²+2e^x(logx)/x)dx

=∫e^x(logx)2dx +∫(2e^x/x)(logx)dx

Integrating by parts,

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

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

= e^x(logx)²+C

Question 18. ∫e^x(sin^-1x+1/(1-x²)^1/2)dx

Solution:

= e^xsin^-1x-∫e^x(d(sin^-1x)/dx) dx +∫e^x/(1-x²)^1/2dx

= e^xsin^-1x- ∫e^x/(1-x²)^1/2dx+∫e^x/(1-x²)^1/2dx

= e^xsin^-1x+C

Question 19. ∫e^2x(-sinx +2cosx)dx

Solution:

= -∫e^2xsinxdx +2∫e^2xcosxdx

= -∫e^2xsinxdx+2((1/2)e^2xcosx+∫(1/2)e^2xsinxdx)

= -∫e^2xsinxdx+e^2xcosx+∫e^2xsinxdx

= e^2xcosx+C

Question 20. ∫e^x(tan^-1x+1/(1+x²))dx

Solution:

= e^xtan^-1x-∫e^x (d(tan^-1x)/dx) dx+∫e^x/(1+x²)dx

= e^xtan^-1x-∫e^x/(1+x²)dx+∫e^x/(1+x²)dx

= e^xtan^-1x+C

Question 21. ∫e^x((sinxcosx-1)/sin²x)dx

Solution:

= ∫e^x(cotx-cosec²x)dx

= e^xcotx-∫e^x(d(cotx)/dx) dx-∫e^xcosec²xdx

= e^xcotx+∫e^xcosec²xdx -∫e^xcosec²xdx

= e^xcotx+C

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

Solution:

Suppose,

logx=z

=>e^z=x

=>d(e^z)/dx=1

=>e^zdz=dx

Substituting it in original question,

∫(tan z+sec²z)e^zdz

= e^ztanz -∫e^z(d(tanz)/dz))dz+∫e^zsec²zdz

= e^ztanz-∫e^zsec²zdz+∫e^zsec²zdz

= e^ztanz+C

= e^(logx)tan(logx)+C

= x tan(logx)+C

Question 23. ∫e^x(x-4)/(x-2)³dx

Solution:

= ∫e^x((x-2)-2)/(x-2)³dx

= ∫e^x((1/(x-2)²)-(2/(x-2)³))dx

= e^x/(x-2)²-∫e^x(d((x-2)^-2)/dx)dx-2∫e^x/(x-2)³dx

= e^x/(x-2)²+2∫e^x/(x-2)³dx -2∫e^x/(x-2)³dx

= e^x/(x-2)²+C

Question 24. ∫e^2x((1-sin2x)/(1-2cosx))dx

Solution:

=∫e^2x((1-sin2x)/2sin²x)dx

=∫e^2x((cosec²x/2)-cotx)dx

Suppose,

I=I_¹+I₂

I₁=1/2∫e^2xcosec²xdx

I₂=-∫e^2xcotxdx

let,

u=e^2x

=du=2e^2xdx

and

∫cosec2xdx=∫dv

=>v=-cotx+C

So,

I₁=1/2[e^2x(-cotx)-∫(-cotx)2e^2xdx]

I₁=1/2(e^2x(-cotx))+∫cotxe^2xdx

Thus,

I=(1/2)(e^2x(-cotx))+∫cotxe^2xdx -∫e^2x cotx dx

=>I=(1/2)(e^2x(-cotx))+C