∫ π 2 π 3 d x tan x − sin x = ∫ π 2 π 3 d x sin x cos x − sin x = ∫ π 2 π 3 cos x d x sin x − sin x cos x = ∫ π 2 π 3 cos x sin x d x sin 2 x ( 1 − cos x ) = ∫ π 2 π 3 cos x sin x d x ( 1 − cos 2 x ) ( 1 − cos x ) = ∫ π 2 π 3 cos x sin x d x ( 1 + cos x ) ( 1 − cos x ) 2 \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}}}} 2 π ∫ 3 π t a n x − s i n x d x = 2 π ∫ 3 π c o s x s i n x − s i n x d x = 2 π ∫ 3 π s i n x − s i n x c o s x c o s x d x = 2 π ∫ 3 π s i n 2 x ( 1 − c o s x ) c o s x s i n x d x = 2 π ∫ 3 π ( 1 − c o s 2 x ) ( 1 − c o s x ) c o s x s i n x d x = 2 π ∫ 3 π ( 1 + c o s x ) ( 1 − c o s x ) 2 c o s x s i n x d x
Let
u = cos x ⇒ d u = − sin x d x , x ∈ [ π 2 ; π 3 ] ⇒ u ∈ [ 0 ; 1 2 ] 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] u = cos x ⇒ d u = − sin x d x , x ∈ [ 2 π ; 3 π ] ⇒ u ∈ [ 0 ; 2 1 ]
Then
∫ π 2 π 3 cos x sin x d x ( 1 + cos x ) ( 1 − cos x ) 2 = ∫ 0 1 2 − u d u ( 1 + u ) ( 1 − u ) 2 \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}}}} 2 π ∫ 3 π ( 1 + c o s x ) ( 1 − c o s x ) 2 c o s x s i n x d x = 0 ∫ 2 1 ( 1 + u ) ( 1 − u ) 2 − u d u
We expand the integrand into partial fractions
− u ( 1 + u ) ( 1 − u ) 2 = A 1 + u + B 1 − u + C ( 1 − u ) 2 = A ( 1 − u ) 2 + B ( 1 − u ) ( 1 + u ) + C ( 1 + u ) ( 1 + u ) ( 1 − u ) 2 = A ( 1 − 2 u + u 2 ) + B ( 1 − u 2 ) + C ( 1 + u ) ( 1 + u ) ( 1 − u ) 2 = u 2 ( A − B ) + u ( − 2 A + C ) + A + B + C ( 1 + u ) ( 1 − u ) 2 \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}}} ( 1 + u ) ( 1 − u ) 2 − u = 1 + u A + 1 − u B + ( 1 − u ) 2 C = ( 1 + u ) ( 1 − u ) 2 A ( 1 − u ) 2 + B ( 1 − u ) ( 1 + u ) + C ( 1 + u ) = ( 1 + u ) ( 1 − u ) 2 A ( 1 − 2 u + u 2 ) + B ( 1 − u 2 ) + C ( 1 + u ) = ( 1 + u ) ( 1 − u ) 2 u 2 ( A − B ) + u ( − 2 A + C ) + A + B + C
A − B = 0 ⇒ A = B A - B = 0 \Rightarrow A = B A − B = 0 ⇒ A = B
− 2 A + C = − 1 ⇒ C = 2 A − 1 - 2A + C = - 1 \Rightarrow C = 2A - 1 − 2 A + C = − 1 ⇒ C = 2 A − 1
A + B + C = 0 ⇒ 2 A + 2 A − 1 = 0 ⇒ A = 1 4 ⇒ B = 1 4 ⇒ C = − 1 2 A + B + C = 0 \Rightarrow 2A + 2A - 1 = 0 \Rightarrow A = \frac{1}{4} \Rightarrow B = \frac{1}{4} \Rightarrow C = - \frac{1}{2} A + B + C = 0 ⇒ 2 A + 2 A − 1 = 0 ⇒ A = 4 1 ⇒ B = 4 1 ⇒ C = − 2 1
Then
∫ 0 1 2 − u d u ( 1 + u ) ( 1 − u ) 2 = 1 4 ∫ 0 1 2 d u 1 + u + 1 4 ∫ 0 1 2 d u 1 − u − 1 2 ∫ 0 1 2 d u ( 1 − u ) 2 = 1 4 ∫ 0 1 2 d ( 1 + u ) 1 + u − 1 4 ∫ 0 1 2 d ( 1 − u ) 1 − u + 1 2 ∫ 0 1 2 d ( 1 − u ) ( 1 − u ) 2 = 1 4 ln ∣ 1 + u ∣ ∣ 0 1 2 − 1 4 ln ∣ 1 − u ∣ ∣ 0 1 2 − 1 2 ( 1 − u ) ∣ 0 1 2 = 1 4 ( ln 3 2 − ln 1 − ln 1 2 + ln 1 ) − 1 2 ( 1 1 2 − 1 1 ) = 1 4 ln 3 − 1 2 \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} 0 ∫ 2 1 ( 1 + u ) ( 1 − u ) 2 − u d u = 4 1 0 ∫ 2 1 1 + u d u + 4 1 0 ∫ 2 1 1 − u d u − 2 1 0 ∫ 2 1 ( 1 − u ) 2 d u = 4 1 0 ∫ 2 1 1 + u d ( 1 + u ) − 4 1 0 ∫ 2 1 1 − u d ( 1 − u ) + 2 1 0 ∫ 2 1 ( 1 − u ) 2 d ( 1 − u ) = 4 1 ln ∣1 + u ∣ ∣ 0 2 1 − 4 1 ln ∣1 − u ∣ ∣ 0 2 1 − 2 ( 1 − u ) 1 ∣ ∣ 0 2 1 = 4 1 ( ln 2 3 − ln 1 − ln 2 1 + ln 1 ) − 2 1 ( 2 1 1 − 1 1 ) = 4 1 ln 3 − 2 1
Answer: 1 4 ln 3 − 1 2 \frac{1}{4}\ln 3 - \frac{1}{2} 4 1 ln 3 − 2 1
#340153 on Dec 2023
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