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

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

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

- Jan 17, 2013

- 1,667

You can useHello

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

1/(1+cos(x))

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

thanks !

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

or you can use that

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

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

[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]

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- Mar 5, 2012

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Since $\cos 2\alpha=2\cos^2\alpha-1$, we have: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 !

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

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.