Question

Evaluate the Integral

\(\int {\frac{{dx}}{{cosx - 1}}} \)\(\)

Short answer

\(\int {\frac{{dx}}{{cosx - 1}}} = - \left( {cosec\,x + cotx} \right) + c\)

When we divide the numerator and denominator by\(\left( {cosx + 1} \right)\), we get the form of\({a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\).so,\(co{s^2}x - 1 = si{n^2}x\).

Then divide both numerator and denominator by\(si{n^2}x\)to simplify the integral.

Step-by-step solution

Step 1:

\(\begin{aligned}{l}I &= \int {\frac{{dx}}{{cosx - 1}}} \\I &= \int {\frac{{\left( {cosx + 1} \right)}}{{\left( {cosx - 1} \right)\left( {cosx + 1} \right)}}dx} \\I &= \int {\frac{{\left( {cosx + 1} \right)}}{{\left( {co{s^2}x - 1} \right)}}dx} \end{aligned}\)

We know,\(1 + si{n^2}x = co{s^2}x\)

So, \(co{s^2}x - 1 = si{n^2}x\)

Substituting \(co{s^2}x - 1 = si{n^2}x\) in \(I\)

\(I = \int {\frac{{\left( {cosx + 1} \right)}}{{si{n^2}x}}} dx\)

Divide both numerator and denominator by \(si{n^2}x\).

\(\begin{aligned}{l}I &= \int {\left( {\frac{{cosx}}{{si{n^2}x}} + \frac{1}{{si{n^2}x}}} \right)} dx\\I &= \int {\frac{{cosx}}{{si{n^2}x}}dx} + \int {\frac{1}{{si{n^2}x}}} dx\end{aligned}\)

We can write \(\frac{{cosx}}{{si{n^2}x}}\) as \(\frac{{cosx}}{{sinx \times sinx}}\),\(\frac{{cosx}}{{sinx}} = cotx\) and \(\frac{1}{{sinx}} = cosecx\).

Therefore, it becomes \(cotx.cosecx\)

Also, we can write \(\frac{1}{{si{n^2}x}} = cose{c^2}x\)

\(I = \int {cotx\,cosecx\,dx + \int {cose{c^2}x\,dx} } \)

We know the integration of \(cose{c^2}x =  - cotx\) and integration of\(cotx\,cosecx =  - cosecx\).

\(\begin{aligned}{l}I &= - cosecx - cotx + c\\I &= - \left( {cosecx + cotx} \right) + c\end{aligned}\)

Hence, \(I = \int {\frac{{dx}}{{cosx - 1}} = } - \left( {cosecx + cotx} \right) + c\)