MHB Integral of 1/(1+cos(x)) Without Substitution Rule

[Yankel]

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]

One could proceed as follows:

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

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

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

 

[alyafey22]

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

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

or you can use that

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

 

[Prove It]

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]

Hello, Yankel!

\int \frac{dx}{1+\cos x}

Multiply by \frac{1-\cos x}{1-\cos x}

. . \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}

. . =\;\frac{1}{\sin^2x} - \frac{1}{\sin x}\,\frac{\cos x}{\sin x} \;=\;\csc^2x - \csc x\cot xTherefore:

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

 

[I like Serena]

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 $$\tan'u=\frac 1 {\cos^2u}$$, 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.