trignometric integrals

**twilightstr** · January 31st 2009, 01:50 PM

integrate integral of dx/(cosx-1).
this is how i tried solving it:
integral dx/(cosx-1)x(cosx+1)/(cosx+1)
=-integral (cosx + 1)/((sinx)^2)
=-integral (cosx)/((sinx)^2) + integral (cscx)^2dx
u=sinx du=cosxdx
=-[integral du/u^2 dx+ integral (1)dx + integral(cotx)^2dx
I don't know how to integrate (cotx)^2.

**wytiaz** · January 31st 2009, 02:21 PM

∫1/(1-cos(x)) dx

Multiply the numerator and the denomintor by (1+cos(x))

= ∫(1+cos(x)) / [(1+cos(x))(1-cos(x))] dx

= ∫(1+cos(x)) / (1-cos²(x)) dx

= ∫(1+cos(x)) / sin²(x)) dx

= ∫(csc²(x) + csc(x)cot(x)) dx

= -cot(x) - csc(x) + c

= -[(cos(x) + 1) / sin(x)] + c

Your start was excellent. It pays to know how to integrate csc²(x).

**Soroban** · January 31st 2009, 02:26 PM

Hello, twilightstr!

You had a good start on it . . .

$\int \frac{dx}{\cos x-1}$

We have: . $-\int\frac{dx}{1-\cos x}$

Multiply by $\frac{1+\cos x}{1+\cos x}\!:\;\;-\int\frac{dx}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x} \;\;=\;\;-\int\frac{(1+\cos x)\,dx}{1 - \cos^2\!x}$

. . $=\;\;-\int\frac{(1+\cos x)\,dx}{\sin^2\!x} \;\;= \;\;-\int\left(\frac{1}{\sin^2\!x} + \frac{\cos x}{\sin^2\!x}\right)\,dx$ . $=\;\;-\int\left(\csc^2\!x + \csc x\cot x\right)\,dx$

. . $=\; -\left(-\cot x - \csc x\right) + C \;\;=\;\;\cot x + \csc x + C$

**red_dog** · January 31st 2009, 02:56 PM

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

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