Câu hỏi:
Tích phân (I = intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {dfrac{{dx}}{{sin x}}} ) có giá trị bằng:
Lời giải tham khảo:
Đáp án đúng: D
Ta có:
(begin{array}{l}I = intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {dfrac{{dx}}{{sin x}}} = intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {dfrac{{sin x}}{{{{sin }^2}x}}} ,dx\ = – intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {dfrac{{dleft( {cos x} right)}}{{1 – {{cos }^2}x}}} \ = – dfrac{1}{2}intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {left( {dfrac{1}{{1 – cos x}} + dfrac{1}{{1 + cos x}}} right)} ;dleft( {cos x} right)\ = dfrac{1}{2}intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {dfrac{1}{{1 – cos x}}dleft( {1 – cos x} right)} – dfrac{1}{2}intlimits_{dfrac{pi }{3}}^{dfrac{pi }{2}} {dfrac{1}{{1 + cos x}}dleft( {1 + cos x} right)} \ = dfrac{1}{2}ln left| {1 – cos x} right|left| {_{dfrac{pi }{3}}^{dfrac{pi }{2}}} right. – dfrac{1}{2}ln left| {1 + cos x} right|left| {_{dfrac{pi }{3}}^{dfrac{pi }{2}}} right.\ = left( {dfrac{1}{2}ln dfrac{1}{2}} right) – dfrac{1}{2}ln dfrac{3}{2} = dfrac{1}{2}ln dfrac{1}{3}end{array})
Chọn đáp án D.
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