Question

Integral of dx/[2cos²x+3cosx]?

Answer 1

`int_a^b 2cos^2x + 3 cosxdx = 0.5sin2x + x + 3sinx + c`

Explanation:

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`

Answer 2

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

Explanation:

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`