1. Class 12
2. Important Question for exams Class 12

Transcript

Ex 7.5, 17 cos﷮𝑥﷯﷮(1− sin﷮𝑥﷯)(2 − sin﷮𝑥)﷯﷯ [Hint: Put sin x = t] Let sin﷮𝑥﷯=𝑡 Diff. 𝑤.𝑟.𝑡.𝑥 𝑐𝑜𝑠﷮𝑥﷯= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮ 𝑐𝑜𝑠﷮𝑥﷯﷯ Now we can write ﷮﷮ cos﷮𝑥﷯﷮(1− sin﷮𝑥﷯)(2 − sin﷮𝑥)﷯﷯﷯𝑑𝑥= ﷮﷮ cos﷮𝑥﷯﷮(1 − 𝑡)(2 − 𝑡) ﷯﷯ 𝑑𝑡﷮ 𝑐𝑜𝑠﷮𝑥﷯﷯ = ﷮﷮ 𝑑𝑡﷮(1 − 𝑡)(2 − 𝑡)﷯﷯ Now we can write 1﷮ 𝑡 − 1﷯ 𝑡 − 2﷯﷯= 𝐴﷮ 𝑡 − 1﷯﷯ + 𝐵﷮ 𝑡 − 2﷯﷯ 1﷮ 𝑡 − 1﷯ 𝑡 − 2﷯﷯= 𝐴 𝑡 − 2﷯ + 𝐵 𝑡 − 1﷯﷮ 𝑡 − 1﷯ 𝑡 − 2﷯﷯ By cancelling denominator 1=𝐴 𝑡 − 2﷯ + 𝐵 𝑡 − 1﷯ Putting t = 1 in (1) 1=𝐴 1−2﷯ + 𝐵 1−1﷯ 1=𝐴 −1﷯ + 𝐵×0 1=−𝐴 𝐴=−1 Similarly putting t = 2 1=𝐴 2−2﷯ + 𝐵 2−1﷯ 1=𝐴×0 + 𝐵 1﷯ 1=𝐵 𝐵=1 Hence we can write 1﷮ 𝑡 − 1﷯ 𝑡 − 2﷯﷯= −1﷮𝑡 − 1﷯ + 1﷮𝑡 − 2﷯ Integrating 𝑤.𝑟.𝑡.𝑥. ﷮﷮ 1﷮ 𝑡 − 1﷯ 𝑡 − 2﷯﷯﷯ 𝑑𝑡= ﷮﷮ −1﷮𝑡 − 1﷯ + 1﷮𝑡 − 2﷯﷯ 𝑑𝑡 =− log ﷮ 𝑡−1﷯﷯+ log﷮ 𝑡−2﷯﷯+𝐶 Putting back t = 𝑠𝑖𝑛﷮𝑥﷯ =− log ﷮ sin﷮𝑥﷯−1﷯﷯+ log﷮ sin﷮𝑥﷯−2﷯﷯+𝐶 = log﷮ sin﷮𝑥﷯−2﷯﷯− log ﷮ sin﷮𝑥﷯−1﷯﷯+𝐶 = log﷮ sin﷮𝑥﷯ − 2﷮ sin﷮𝑥﷯ − 1﷯﷯﷯+𝐶 = log﷮ −1﷯ 2 − sin﷮𝑥﷯﷯﷮ −1﷯ 1 − sin﷮𝑥﷯﷯﷯﷯﷯+𝐶 = 𝒍𝒐𝒈﷮ 𝟐 − 𝒔𝒊𝒏﷮𝒙﷯﷮𝟏 − 𝒔𝒊𝒏﷮𝒙﷯﷯﷯﷯+𝑪

Class 12
Important Question for exams Class 12