# Integral of dx/[2cos²x+3cosx]?

Jun 10, 2018

${\int}_{a}^{b} 2 {\cos}^{2} x + 3 \cos x \mathrm{dx} = 0.5 \sin 2 x + x + 3 \sin x + c$

#### Explanation:

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

$\cos 2 x = 2 {\cos}^{2} x - 1$
$2 {\cos}^{2} x = \cos 2 x + 1$
${\cos}^{2} x = 0.5 \cos 2 x + 0.5$

Substitute this in to your integral:

${\int}_{a}^{b} 2 \left(0.5 \cos 2 x + 0.5\right) + 3 \cos x \mathrm{dx}$
$= {\int}_{a}^{b} \cos 2 x + 1 + 3 \cos x \mathrm{dx}$

Now you can integrate each term separately:

${\int}_{a}^{b} \cos 2 x + 1 + 3 \cos x \mathrm{dx}$
$= 0.5 \sin 2 x + x + 3 \sin x + c$

Jun 10, 2018

$I = \frac{1}{3} \ln | \sec x + \tan x | - \frac{4}{3 \sqrt{5}} {\tan}^{-} 1 \left(\tan \frac{\frac{x}{2}}{\sqrt{5}}\right) + C$

#### Explanation:

Here,

I=intdx/[2cos²x+3cosx]

$= \int \frac{1}{\cos x \left(2 \cos x + 3\right)} \mathrm{dx}$

$= \frac{1}{3} \int \frac{3}{\cos x \left(2 \cos x + 3\right)} \mathrm{dx}$

$= \frac{1}{3} \int \frac{\left(2 \cos x + 3\right) - 2 \cos x}{\cos x \left(2 \cos x + 3\right)} \mathrm{dx}$

$= \frac{1}{3} \int \left[\frac{1}{\cos} x - \frac{2}{2 \cos x + 3}\right] \mathrm{dx}$

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

$= \frac{1}{3} \textcolor{red}{\ln | \sec x + \tan x |} - \frac{2}{3} {I}_{1.} . . \to \left(A\right)$

Now, ${I}_{1} = \int \frac{1}{2 \cos x + 3} \mathrm{dx}$

Let, $\textcolor{v i o \le t}{\tan \left(\frac{x}{2}\right) = t} \implies {\sec}^{2} \left(\frac{x}{2}\right) \frac{1}{2} \mathrm{dx} = \mathrm{dt}$

$\implies \left(1 + {\tan}^{2} \left(\frac{x}{2}\right)\right) \mathrm{dx} = 2 \mathrm{dt} \implies \mathrm{dx} = \frac{2 \mathrm{dt}}{1 + {t}^{2}}$

So,

${I}_{1} = \int \frac{2}{2 \left(\frac{1 - {t}^{2}}{1 + {t}^{2}}\right) + 3} \times \frac{1}{1 + {t}^{2}} \mathrm{dt}$

$= 2 \int \frac{1}{2 - 2 {t}^{2} + 3 + 3 {t}^{2}} \mathrm{dt}$

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

$= 2 \textcolor{b l u e}{\left[\frac{1}{\sqrt{5}} {\tan}^{-} 1 \left(\frac{t}{\sqrt{5}}\right)\right]} + c$

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

${I}_{1} = \frac{2}{\sqrt{5}} {\tan}^{-} 1 \left(\tan \frac{\frac{x}{2}}{\sqrt{5}}\right) + c$

From $\left(A\right)$ ,we get

$I = \frac{1}{3} \ln | \sec x + \tan x | - \frac{2}{3} \times \frac{2}{\sqrt{5}} {\tan}^{-} 1 \left(\tan \frac{\frac{x}{2}}{\sqrt{5}}\right) + C$

$I = \frac{1}{3} \ln | \sec x + \tan x | - \frac{4}{3 \sqrt{5}} {\tan}^{-} 1 \left(\tan \frac{\frac{x}{2}}{\sqrt{5}}\right) + C$

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