# Thread: Integrate 1/[5+4cos(x)] with limits?

1. ## Integrate 1/[5+4cos(x)] with limits?

I also have limits - zero and Pi.
I seem to have obtained an expression for the integral but substituting Pi makes it troublesome.
so, (apologies for the lack of LaTeX)
let u = tan(x/2)

u = sin(x/2) / cos(x/2)
u cos(x/2) = sin(x/2)
u^2 cos^2(x/2) = sin^2(x/2)
u^2 cos^2(x/2) = (1 - cos^2(x/2)
cos^2(x/2)(1 + u^2) = 1
cos^2(x/2) = 1/ (1 + u^2)
sin^2(x/2) = 1 - 1 /(1 + u^2) = u^2/(1 + u^2)
cos x = cos^2(x/2) - sin^2(x/2)
= 1 / (1 + u^2) - u^2/(1 + u^2) = (1 - u^2)/(1 + u^2)
cos x = (1 - u^2)/(1 + u^2)
when u = tan(x/2)
x/2 = arctan(u)
dx = 2du / (1 + u^2)

substituting these into integral

∫ dx /(5 + 4 cos x)

= ∫ 2du /(1 + u^2)(5 + 4(1 - u^2)/(1 + u^2)

= ∫ 2du/ (5 + 5u^2 + 4 - 4u^2)

= ∫ 2du/ (9 + u^2)

= 2/9 ∫du/ [1 + (u/3)^2)]

= (2/3) ∫(du/3)/ [1 + (u/3)^2)]

= (2/3) arctan(u/3) + C

substitute u/3 = (1/3)tan(x/2)

= (2/3) arctan[1/3tan(x/2) ] + C

but putting Pi into tan(x/2) makes the expression undefined! where did i go wrong?

thank you

2. Let $u=\tan(\frac{x}{2})$, then $du=\frac{1}{2}\sec^2(\frac{x}{2})dx$

Now use the substitutions $\sin(x)=\frac{2u}{u^2+1}$, $\cos(x)=\frac{1-u^2}{u^2+1}$, and $dx=\frac{2du}{u^2+1}$. You'll find that some really good simplification happens.

4. It's too hard to read. Check out this link: Learn Latex so we don't go crazy looking at that stuff.

So your integrand Looks like this: $\frac{2}{(u^2+1)\left[\frac{4(1-u^2)}{u^2+1}+4\right]}$

5. Hello, cassius1

Your work is correct . . .

Note that: . $\tan\frac{\pi}{2} \:=\:\infty$

So we may write: . $\arctan(\infty) \:=\:\frac{\pi}{2}$

We have: . $\displaystyle \int^{\pi}_0 \frac{dx}{5 + 4\cos x} \;=\;\tfrac{2}{3}\arctan\left(\tfrac{1}{3}\tan\tfr ac{x}{2}\right)\,\bigg]^{\pi}_0$

. . . . . . $=\;\frac{2}{3}\arctan\left(\frac{1}{3}\tan\frac{\p i}{2}\right) - \frac{2}{3}\arctan\left(\frac{1}{3}\tan 0\right)$

. . . . . . $=\;\frac{2}{3}\underbrace{\arctan(\infty)} - \frac{2}{3}\arctan(0)$

. . . . . . $=\quad\;\,\dfrac{2}{3}\left(\dfrac{\pi}{2}\right) \quad-\quad 0$

. . . . . . $=\qquad\;\;\dfrac{\pi}{3}$

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