Question

Integration by Partial Fractions

∫dx/(tanx-sinx) from π/2 to π/3 

Answer 1

$\int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{dx}}{{\tan x - \sin x}}} = \int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{dx}}{{\frac{{\sin x}}{{\cos x}} - \sin x}}} = \int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{\cos xdx}}{{\sin x - \sin x\cos x}} = } \int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{\cos x\sin xdx}}{{{{\sin }^2}x\left( {1 - \cos x} \right)}}} = \int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{\cos x\sin xdx}}{{\left( {1 - {{\cos }^2}x} \right)\left( {1 - \cos x} \right)}}} = \int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{\cos x\sin xdx}}{{\left( {1 + \cos x} \right){{\left( {1 - \cos x} \right)}^2}}}}$

Let

$u = \cos x \Rightarrow du = - \sin xdx,\,\,x \in \left[ {\frac{\pi }{2};\,\frac{\pi }{3}} \right] \Rightarrow u \in \left[ {0;\,\frac{1}{2}} \right]$

Then

$\int\limits_{\frac{\pi }{2}}^{\frac{\pi }{3}} {\frac{{\cos x\sin xdx}}{{\left( {1 + \cos x} \right){{\left( {1 - \cos x} \right)}^2}}}} = \int\limits_0^{\frac{1}{2}} {\frac{{ - udu}}{{\left( {1 + u} \right){{\left( {1 - u} \right)}^2}}}}$

We expand the integrand into partial fractions

$\frac{{ - u}}{{\left( {1 + u} \right){{\left( {1 - u} \right)}^2}}} = \frac{A}{{1 + u}} + \frac{B}{{1 - u}} + \frac{C}{{{{\left( {1 - u} \right)}^2}}} = \frac{{A{{\left( {1 - u} \right)}^2} + B(1 - u)(1 + u) + C(1 + u)}}{{\left( {1 + u} \right){{\left( {1 - u} \right)}^2}}} = \frac{{A\left( {1 - 2u + {u^2}} \right) + B\left( {1 - {u^2}} \right) + C(1 + u)}}{{\left( {1 + u} \right){{\left( {1 - u} \right)}^2}}} = \frac{{{u^2}(A - B) + u( - 2A + C) + A + B + C}}{{\left( {1 + u} \right){{\left( {1 - u} \right)}^2}}}$

$A - B = 0 \Rightarrow A = B$

$- 2A + C = - 1 \Rightarrow C = 2A - 1$

$A + B + C = 0 \Rightarrow 2A + 2A - 1 = 0 \Rightarrow A = \frac{1}{4} \Rightarrow B = \frac{1}{4} \Rightarrow C = - \frac{1}{2}$

Then

$\int\limits_0^{\frac{1}{2}} {\frac{{ - udu}}{{\left( {1 + u} \right){{\left( {1 - u} \right)}^2}}}} = \frac{1}{4}\int\limits_0^{\frac{1}{2}} {\frac{{du}}{{1 + u}}} + \frac{1}{4}\int\limits_0^{\frac{1}{2}} {\frac{{du}}{{1 - u}}} - \frac{1}{2}\int\limits_0^{\frac{1}{2}} {\frac{{du}}{{{{\left( {1 - u} \right)}^2}}}} = \frac{1}{4}\int\limits_0^{\frac{1}{2}} {\frac{{d\left( {1 + u} \right)}}{{1 + u}}} - \frac{1}{4}\int\limits_0^{\frac{1}{2}} {\frac{{d\left( {1 - u} \right)}}{{1 - u}}} + \frac{1}{2}\int\limits_0^{\frac{1}{2}} {\frac{{d\left( {1 - u} \right)}}{{{{\left( {1 - u} \right)}^2}}}} = \frac{1}{4}\ln \left. {|1 + u|} \right|_0^{\frac{1}{2}} - \frac{1}{4}\ln \left. {|1 - u|} \right|_0^{\frac{1}{2}} - \left. {\frac{1}{{2\left( {1 - u} \right)}}} \right|_0^{\frac{1}{2}} = \frac{1}{4}\left( {\ln \frac{3}{2} - \ln 1 - \ln \frac{1}{2} + \ln 1} \right) - \frac{1}{2}\left( {\frac{1}{{\frac{1}{2}}} - \frac{1}{1}} \right) = \frac{1}{4}\ln 3 - \frac{1}{2}$

Answer: $\frac{1}{4}\ln 3 - \frac{1}{2}$