Integral of dx/[2cos²x+3cosx]?

Before you integrate this you need to convert the #cos^2x# in to something that you can integrate. This requires use of double angle formulae:

#cos2x = 2cos^2x - 1#
#2cos^2x = cos2x + 1#
#cos^2x = 0.5cos2x + 0.5#

Substitute this in to your integral:

#int_a^b 2(0.5cos2x + 0.5) + 3 cosxdx#
#= int_a^b cos2x + 1 + 3 cosxdx#

Now you can integrate each term separately:

#int_a^b cos2x + 1 + 3 cosxdx#
#= 0.5sin2x + x + 3sinx + c#

Here,

#I=intdx/[2cos²x+3cosx]#

#=int1/(cosx(2cosx+3))dx#

#=1/3int3/(cosx(2cosx+3))dx#

#=1/3int[(2cosx+3)-2cosx)/(cosx(2cosx+3))dx#

#=1/3int[1/cosx-2/(2cosx+3)]dx#

#=1/3color(red)(intsecxdx)-2/3int1/(2cosx+3)dx...tocolor(red)(Apply(1)#

#=1/3color(red)(ln|secx+tanx|)-2/3I_1...to(A)#

Now, #I_1=int1/(2cosx+3)dx#

Let, #color(violet)(tan(x/2)=t)=>sec^2(x/2)1/2dx=dt#

#=>(1+tan^2(x/2))dx=2dt=>dx=(2dt)/(1+t^2)#

So,

#I_1=int2/(2((1-t^2)/(1+t^2))+3)xx1/(1+t^2)dt#

#=2int1/(2-2t^2+3+3t^2)dt#

#=2intcolor(blue)(1/(t^2+(sqrt5)^2)dt...tocolor(blue)(Apply(2)#

#=2color(blue)([1/sqrt5tan^-1(t/sqrt5)])+c#

Subst. back ,#color(violet)(t=tan(x/2)#

#I_1=2/sqrt5tan^-1(tan(x/2)/sqrt5)+c#

From #(A)# ,we get

#I=1/3ln|secx+tanx|-2/3xx2/sqrt5tan^-1(tan(x/2)/sqrt5)+C#

#I=1/3ln|secx+tanx|-4/(3sqrt5)tan^-1(tan(x/2)/sqrt5)+C#

Note : Formulas.

#color(red)((1)intsecxdx=ln|secx+tanx|+c#

#color(blue)((2)int1/(x^2+a^2)dx=1/atan^-1(x/a)+c#