Ex 7.3, 14 - Chapter 7 Class 12 Integrals (Term 2)

Transcript

Ex 7.3, 14 Integrate the function cos⁡〖𝑥 − sin⁡𝑥 〗/(1 + sin⁡2𝑥 ) ∫1▒cos⁡〖𝑥 − sin⁡𝑥 〗/(1 + sin⁡2𝑥 ) 𝑑𝑥 =∫1▒cos⁡〖𝑥 −〖 sin〗⁡𝑥 〗/(𝟏 + 2 sin⁡𝑥 cos⁡𝑥 ) 𝑑𝑥 =∫1▒cos⁡〖𝑥 −〖 sin〗⁡𝑥 〗/(〖𝐬𝐢𝐧〗^𝟐⁡𝒙 + 〖𝐜𝐨𝐬〗^𝟐⁡𝒙 + 2 sin⁡cos⁡𝑥 ) 𝑑𝑥 =∫1▒cos⁡〖𝑥 −〖 sin〗⁡𝑥 〗/(sin⁡𝑥 + cos⁡𝑥 )^2 𝑑𝑥 Let sin⁡𝑥+cos⁡𝑥=𝑡 Differentiating w.r.t.x (𝑠𝑖𝑛⁡2 𝜃=2 𝑠𝑖𝑛⁡𝜃 𝑐𝑜𝑠⁡𝜃) (As 〖𝑠𝑖𝑛〗^2⁡𝜃+〖𝑐𝑜𝑠〗^2⁡𝜃=1) 𝑑(sin⁡𝑥 + cos⁡𝑥 )/𝑑𝑥=𝑑𝑡/𝑑𝑥 cos 𝑥−sin⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(cos 𝑥 − sin⁡𝑥 ) Thus, our equation becomes ∫1▒((cos⁡〖𝑥 − sin⁡𝑥 〗 ))/(sin⁡𝑥 + cos⁡𝑥 )^2 𝑑𝑥 =∫1▒((cos⁡〖𝑥 − sin⁡𝑥 〗 ))/𝑡^2 ×𝑑𝑡/cos⁡〖𝑥 − sin⁡𝑥 〗 =∫1▒𝑑𝑡/𝑡^2 =∫1▒𝑡^(−2) 𝑑𝑡 =𝑡^(−2 +1)/(−2 + 1) +𝐶 =𝑡^(−1)/(−1) +𝐶 =(−1)/𝑡 +𝐶 Putting value of 𝑡=𝑠𝑖𝑛⁡𝑥+𝑐𝑜𝑠 𝑥 =(−𝟏)/(𝐬𝐢𝐧⁡𝒙 + 𝒄𝒐𝒔⁡𝒙 ) +𝑪