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Example 30 Evaluate ∫_0^πœ‹β–’(π‘₯ 𝑠𝑖𝑛 π‘₯)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯ Let I=∫_0^πœ‹β–’(π‘₯ sin⁑π‘₯)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯ ∴ I=∫_0^πœ‹β–’((πœ‹ βˆ’ π‘₯) 𝑠𝑖𝑛(πœ‹ βˆ’ π‘₯))/(1 + cos^2⁑(πœ‹ βˆ’ π‘₯) ) 𝑑π‘₯ I=∫_0^πœ‹β–’γ€–((πœ‹ βˆ’ π‘₯) sin⁑π‘₯)/(1 + [βˆ’cos⁑π‘₯ ]^2 ) 𝑑π‘₯γ€— I=∫_0^πœ‹β–’γ€–(πœ‹ sin⁑〖π‘₯ βˆ’ π‘₯ sin⁑π‘₯ γ€—)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯γ€— Adding (1) and (2) i.e. (1) + (2) I +I=∫_0^πœ‹β–’(π‘₯ sin⁑π‘₯)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯+∫_0^πœ‹β–’(πœ‹ sin⁑〖π‘₯ βˆ’ π‘₯ 𝑠𝑖𝑛π‘₯γ€—)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯ 2I =∫_0^πœ‹β–’(π‘₯ sin⁑〖π‘₯ + πœ‹ sin⁑〖π‘₯ βˆ’ π‘₯ sin⁑π‘₯ γ€— γ€—)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯ 2I =∫_0^πœ‹β–’(πœ‹ sin⁑〖π‘₯ γ€—)/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯ 2I =πœ‹βˆ«_0^πœ‹β–’sin⁑〖π‘₯ γ€—/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯ ∴ I = πœ‹/2 ∫_0^πœ‹β–’γ€–sin⁑π‘₯/(1 + cos^2⁑π‘₯ ) 𝑑π‘₯γ€— Let cos π‘₯ = t Differentiate both sides w.r.t.π‘₯ – sin⁑〖π‘₯=𝑑𝑑/𝑑π‘₯γ€— 𝑑π‘₯=𝑑𝑑/(βˆ’sin⁑π‘₯ ) Putting the values of (cos⁑π‘₯ ) and dπ‘₯, we get I =πœ‹/2 ∫1_0^πœ‹β–’γ€–sin⁑π‘₯/(1 + 𝑑^2 ).𝑑π‘₯γ€— I = πœ‹/2 ∫1_0^πœ‹β–’γ€–sin⁑π‘₯/(1 + 𝑑^2 ) Γ— 𝑑𝑑/(βˆ’sin⁑π‘₯ )γ€— I = (βˆ’πœ‹)/2 ∫1_0^πœ‹β–’γ€–1/(1 + 𝑑^2 ) .𝑑𝑑〗 I = (βˆ’πœ‹)/2 ∫1_0^πœ‹β–’γ€–1/((1)^2 + (𝑑)^2 ) .𝑑𝑑〗 I=(βˆ’πœ‹)/2 [1/1 tan^(βˆ’1)⁑(𝑑/1) ]_π‘œ^πœ‹ =(βˆ’πœ‹)/2 [tan^(βˆ’1)⁑(𝑑) ]_0^πœ‹ Putting t = cos x =(βˆ’πœ‹)/2 [tan^(βˆ’1)⁑(cos⁑π‘₯ ) ]_0^πœ‹ =(βˆ’πœ‹)/2 [tan^(βˆ’1)⁑〖[cosβ‘πœ‹ ]βˆ’tan^(βˆ’1)⁑[cos⁑0 ] γ€— ] =(βˆ’πœ‹)/2 [tan^(βˆ’1)⁑〖(βˆ’1)βˆ’tan^(βˆ’1)⁑(1) γ€— ] =(βˆ’πœ‹)/2 [βˆ’tan^(βˆ’1)⁑〖(1)βˆ’tan^(βˆ’1)⁑(1) γ€— ] =(βˆ’πœ‹)/2 [βˆ’2 tan^(βˆ’1)⁑(1) ] =πœ‹[tan^(βˆ’1)⁑(1) ] = πœ‹[πœ‹/4] =𝝅^𝟐/πŸ’

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.