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Trigonometric Integral

Yankel

Active member
Jan 27, 2012
398
Hello

I am trying to solve the integral of the next function:

1/(1+cos(x))

and without using the substitution rule....any ideas ?

thanks !
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
One could proceed as follows:

\(\displaystyle \frac{1}{\cos(x)+1}\cdot\frac{\cos(x)+1}{\cos(x)+1}=\frac{\cos(x)+1}{(\cos(x)+1)^2}= \frac{\cos(x)+\sin^2(x)+\cos^2(x)}{(\cos(x)+1)^2}=\)

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

\(\displaystyle \frac{d}{dx}\left(\frac{\sin(x)}{\cos(x)+1} \right)\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Hello

I am trying to solve the integral of the next function:

1/(1+cos(x))

and without using the substitution rule....any ideas ?

thanks !
You can use

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

or you can use that

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hello

I am trying to solve the integral of the next function:

1/(1+cos(x))

and without using the substitution rule....any ideas ?

thanks !
You will have a LOT of difficulty trying to integrate this function without making a substitution.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Yankel!

[tex]\int \frac{dx}{1+\cos x}[/tex]

Multiply by [tex]\frac{1-\cos x}{1-\cos x}[/tex]

. . [tex]\frac{1}{1+\cos x} \cdot\frac{1-\cos x}{1-\cos x} \;=\;\frac{1-\cos x}{1-\cos^2x} \;=\;\frac{1-\cos x}{\sin^2x}[/tex]

. . [tex]=\;\frac{1}{\sin^2x} - \frac{1}{\sin x}\,\frac{\cos x}{\sin x} \;=\;\csc^2x - \csc x\cot x [/tex]


Therefore:

.. [tex]\int(\csc^2x - \csc x\cot x)\,dx \;=\; -\cot x + \csc x + C[/tex]
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,855
Hello

I am trying to solve the integral of the next function:

1/(1+cos(x))

and without using the substitution rule....any ideas ?

thanks !
Since $\cos 2\alpha=2\cos^2\alpha-1$, we have:
$$\frac 1 {1+cos(x)} = \frac 1 {1+(2\cos^2 (x/2)-1)} = \frac 1 {\cos^2(x/2)} \cdot \frac 1 2$$

Since \(\displaystyle \tan'u=\frac 1 {\cos^2u}\), it follows that
$$\int \frac {dx} {1+cos(x)} = \int \frac 1 {\cos^2(x/2)} \cdot \frac 1 2 dx = \tan \frac x 2 + C$$
Verify by applying the chain rule to the right hand side.