# Another Tricky Integral

• Sep 8th 2008, 09:32 PM
RedBarchetta
Another Tricky Integral
$\displaystyle \int {\frac{1} {{1 + \cos x}}dx = \tan \tfrac{x} {2} + C}$

Alright. Would I multiply the bottom part by [1-cos(x)]?

Here's what I tried:

$\displaystyle \int {\frac{{1(1 - \cos x)}} {{(1 + \cos x)(1 - \cos x)}}dx} = \int {\frac{{1 - \cos x}} {{1 - \cos ^2 x}}dx}$

Now I split these up into two integrals:

$\displaystyle \begin{gathered} \sin ^2 x + \cos ^2 x = 1 \Rightarrow \sin ^2 x = 1 - \cos ^2 x \hfill \\ \int {\frac{1} {{\sin ^2 x}}dx} - \int {\frac{{\cos x}} {{\sin ^2 x}}dx} \hfill \\ \int {\csc ^2 xdx} - \int {\cot x\csc xdx} = \csc x - \cot x + c \hfill \\ \end{gathered}$

....am I right?.....or where did I go wrong.

Thank you.
• Sep 8th 2008, 09:47 PM
o_O
It can be shown that: $\displaystyle \tan \left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \csc x - \cot x$

Basically, yeah you're good (Yes)
• Sep 8th 2008, 09:48 PM
Moo
Hello,
Quote:

Originally Posted by RedBarchetta
$\displaystyle \int {\frac{1} {{1 + \cos x}}dx = \tan \tfrac{x} {2} + C}$

Alright. Would I multiply the bottom part by [1-cos(x)]?

(Clapping)

Quote:

Here's what I tried:

$\displaystyle \int {\frac{{1(1 - \cos x)}} {{(1 + \cos x)(1 - \cos x)}}dx} = \int {\frac{{1 - \cos x}} {{1 - \cos ^2 x}}dx}$

Now I split these up into two integrals:

$\displaystyle \begin{gathered} \sin ^2 x + \cos ^2 x = 1 \Rightarrow \sin ^2 x = 1 - \cos ^2 x \hfill \\ \int {\frac{1} {{\sin ^2 x}}dx} - \int {\frac{{\cos x}} {{\sin ^2 x}}dx} \hfill \\ \int {\csc ^2 xdx} - \int {\cot x\csc xdx} = \csc x - \cot x + c \hfill \\ \end{gathered}$

....am I right?.....or where did I go wrong.

Thank you.

if you want to transform it into tan(x/2) :

$\displaystyle \frac 1{\sin(x)}-\frac{\cos(x)}{\sin(x)}=\frac{1-\cos(x)}{\sin(x)}=\frac{1-(1-2\sin^2 \tfrac x2)}{2 \cos \tfrac x2 \sin \tfrac x2}$

$\displaystyle =\frac{2 \sin^2 \tfrac x2}{2 \cos \tfrac x2 \sin \tfrac x2}$

Is there anything you can do ? (Tongueout)
• Sep 8th 2008, 09:54 PM
mr fantastic
Quote:

Originally Posted by RedBarchetta
$\displaystyle \int {\frac{1} {{1 + \cos x}}dx = \tan \tfrac{x} {2} + C}$

Alright. Would I multiply the bottom part by [1-cos(x)]?

Here's what I tried:

$\displaystyle \int {\frac{{1(1 - \cos x)}} {{(1 + \cos x)(1 - \cos x)}}dx} = \int {\frac{{1 - \cos x}} {{1 - \cos ^2 x}}dx}$

Now I split these up into two integrals:

$\displaystyle \begin{gathered} \sin ^2 x + \cos ^2 x = 1 \Rightarrow \sin ^2 x = 1 - \cos ^2 x \hfill \\ \int {\frac{1} {{\sin ^2 x}}dx} - \int {\frac{{\cos x}} {{\sin ^2 x}}dx} \hfill \\ \int {\csc ^2 xdx} - \int {\cot x\csc xdx} = \csc x - \cot x + c \hfill \\ \end{gathered}$

....am I right?.....or where did I go wrong.

Thank you.

The easier way:

$\displaystyle 1 + \cos x = 1 + \cos^2 \left( \frac{x}{2} \right) - \sin^2 \left( \frac{x}{2} \right)$

using the double angle formula

$\displaystyle = 2 \cos^2 \left( \frac{x}{2} \right)$.

Therefore your integral becomes $\displaystyle \int \frac{1}{2 \cos^2 \left( \frac{x}{2} \right)} \, dx = \frac{1}{2} \int \sec^2 \left( \frac{x}{2} \right) \, dx = \tan \left( \frac{x}{2} \right) + C$.