Check sibling questions

Ex 7.3, 14 - Integrate cos x - sin x / 1 + sin 2x - Teachoo

Ex 7.3, 14 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.3, 14 - Chapter 7 Class 12 Integrals - Part 3


Transcript

Ex 7.3, 14 Integrate the function cos⁑〖π‘₯ βˆ’ sin⁑π‘₯ γ€—/(1 + sin⁑2π‘₯ ) ∫1β–’cos⁑〖π‘₯ βˆ’ sin⁑π‘₯ γ€—/(1 + sin⁑2π‘₯ ) 𝑑π‘₯ =∫1β–’cos⁑〖π‘₯ βˆ’γ€– sin〗⁑π‘₯ γ€—/(𝟏 + 2 sin⁑π‘₯ cos⁑π‘₯ ) 𝑑π‘₯ =∫1β–’cos⁑〖π‘₯ βˆ’γ€– sin〗⁑π‘₯ γ€—/(〖𝐬𝐒𝐧〗^πŸβ‘π’™ + γ€–πœπ¨π¬γ€—^πŸβ‘π’™ + 2 sin⁑cos⁑π‘₯ ) 𝑑π‘₯ =∫1β–’cos⁑〖π‘₯ βˆ’γ€– sin〗⁑π‘₯ γ€—/(sin⁑π‘₯ + cos⁑π‘₯ )^2 𝑑π‘₯ Let sin⁑π‘₯+cos⁑π‘₯=𝑑 Differentiating w.r.t.x (𝑠𝑖𝑛⁑2 πœƒ=2 π‘ π‘–π‘›β‘πœƒ π‘π‘œπ‘ β‘πœƒ) (As 〖𝑠𝑖𝑛〗^2β‘πœƒ+γ€–π‘π‘œπ‘ γ€—^2β‘πœƒ=1) 𝑑(sin⁑π‘₯ + cos⁑π‘₯ )/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ cos π‘₯βˆ’sin⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(cos π‘₯ βˆ’ sin⁑π‘₯ ) Thus, our equation becomes ∫1β–’((cos⁑〖π‘₯ βˆ’ sin⁑π‘₯ γ€— ))/(sin⁑π‘₯ + cos⁑π‘₯ )^2 𝑑π‘₯ =∫1β–’((cos⁑〖π‘₯ βˆ’ sin⁑π‘₯ γ€— ))/𝑑^2 ×𝑑𝑑/cos⁑〖π‘₯ βˆ’ sin⁑π‘₯ γ€— =∫1▒𝑑𝑑/𝑑^2 =∫1▒𝑑^(βˆ’2) 𝑑𝑑 =𝑑^(βˆ’2 +1)/(βˆ’2 + 1) +𝐢 =𝑑^(βˆ’1)/(βˆ’1) +𝐢 =(βˆ’1)/𝑑 +𝐢 Putting value of 𝑑=𝑠𝑖𝑛⁑π‘₯+π‘π‘œπ‘  π‘₯ =(βˆ’πŸ)/(𝐬𝐒𝐧⁑𝒙 + 𝒄𝒐𝒔⁑𝒙 ) +π‘ͺ

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.