# Another Tricky Integral

• September 8th 2008, 10:32 PM
RedBarchetta
Another Tricky Integral
$
\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:

$
\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:

$
\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.
• September 8th 2008, 10:47 PM
o_O
It can be shown that: $\tan \left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \csc x - \cot x$

Basically, yeah you're good (Yes)
• September 8th 2008, 10:48 PM
Moo
Hello,
Quote:

Originally Posted by RedBarchetta
$
\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:

$
\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:

$
\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) :

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

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

Is there anything you can do ? (Tongueout)
• September 8th 2008, 10:54 PM
mr fantastic
Quote:

Originally Posted by RedBarchetta
$
\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:

$
\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:

$
\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:

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

using the double angle formula

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

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