Alternatives for evaluating $ \int \frac { 1 } { 5 + 4 \cos x} \ dx $ ??

Question

$$ \int \frac { 1 } { 5 + 4 \cos x} \ dx $$

$$ \text{The solution given in the book for solving this was to use the identity.} $$

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

I was wondering if there was any other way for solving this ?

Answer 1

You can also solve it using the substitution $x=2\arctan t$ and $\mathrm dx=\frac2{1+t^2}\,\mathrm dt$. This will transform your problem into the computation of a primitive of a rational function.

Answer 2

$$I=\int\frac{1}{5+4\cos(x)}dx$$ now we can use Weierstrauss substitution: $$u=\tan\left(\frac{x}{2}\right)$$ so: $$dx=\frac{2}{1+u^2}du$$ and $$\cos(x)=\frac{1-u^2}{1+u^2}$$ now we can transform our integral to: $$I=\int\frac{1}{5+4\frac{1-u^2}{1+u^2}}\frac{2}{1+u^2}du=\int\frac{1}{5(1+u^2)+4(1-u^2)}du=\int\frac{1}{9+u^2}du=\frac{1}{9}\int\frac{1}{1+\left(\frac{u}{3}\right)^2}du$$ now let:$$u=3\tan(v)$$ $$du=3\sec^2(v)dv$$ so: $$I=\frac{1}{9}\int3\frac{\sec^2(v)}{1+ \tan^2(v)}dv=\frac{1}{3}\int dv=\frac{v}{3}+C=\frac{\arctan\left(\frac{u}{3}\right)}{3}+C=\frac{\arctan\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right)}{3}+C$$

Answer 3

Cheating a little, $$x=2 \cot ^{-1}\left(\frac{u}{3}\right)\implies dx=-\frac{6}{u^2+9}\,du$$ $$\int \frac { dx } { 5 + 4 \cos x} =-\frac 23\int \frac {du}{1+u^2}$$

Answer 4

$$\int\frac{1}{5+4\cos x}dx=\int\frac{1}{5+4\frac{e^{ix}+e^{-ix}}{2}} \ dx= \int\frac{2e^{ix}}{4e^{2ix}+10e^{ix}+4} \ dx $$ let $u = e^{ix}$ and $du = ie^{ix}$ $$\int\frac{-2i}{2u^2+10u+4}\ du = -i\int\frac{1}{(u+\frac{1}{2})(u+2)}du$$ and the rest is easy!

Answer 5

Let $$\displaystyle I =\int\frac{1}{5+4\cos x}dx=\int\frac{5-4\cos x}{25-16\cos^2 x}dx$$

$$ I =5\int\frac{(\tan x)'}{9+16\tan^2 x}dx-4\int\frac{(\sin x)'}{25-16\cos^2 x}dx$$