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Misc 39 Choose the correct answer ∫▒cos⁡2𝑥/((sin⁡𝑥 + cos⁡𝑥 )^2 ) 𝑑𝑥 is equal to (A) (−1)/(sin⁡𝑥 + cos⁡𝑥 )+𝐶 (B) log⁡|sin⁡𝑥+cos⁡𝑥 |+𝐶 (C) log⁡|sin⁡𝑥−cos⁡𝑥 |+𝐶 (D) " " 1/(sin⁡𝑥 + cos⁡𝑥 )^2 ∫1▒cos⁡2𝑥/(cos⁡𝑥 + sin⁡𝑥 )^2 =∫1▒(cos^2⁡𝑥 − sin^2⁡𝑥)/(cos⁡𝑥 + sin⁡𝑥 )^2 𝑑𝑥 =∫1▒(cos⁡𝑥 − sin⁡𝑥 )(cos⁡𝑥 + sin⁡𝑥 )/(cos⁡𝑥 + sin⁡𝑥 )^2 𝑑𝑥 =∫1▒(cos⁡𝑥 − sin⁡𝑥)/(cos⁡𝑥 + sin⁡𝑥 ) 𝑑𝑥 Let cos⁡𝑥+sin⁡𝑥=𝑡 Diff w.r.t. x −sin⁡𝑥+cos⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=1/((cos⁡𝑥 − sin⁡𝑥 ) ) 𝑑𝑡 Hence, our equation becomes =∫1▒((cos⁡𝑥 − sin⁡𝑥 ))/𝑑𝑡 ×𝑑𝑡/(cos⁡𝑥 − sin⁡𝑥 ) =∫1▒1/𝑡 𝑑𝑡 =log⁡𝑡+𝐶 Putting value of 𝑡=𝑐𝑜𝑠⁡𝑥+𝑠𝑖𝑛⁡𝑥 =log⁡|cos⁡𝑥+sin⁡𝑥 |+𝐶 Hence correct answer is B.

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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.