Ex 7.2, 21 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.2, 21 tan﷮2﷯ (2𝑥 – 3) Let I = ﷮﷮ tan﷮2﷯ (2𝑥 – 3)﷯ . 𝑑𝑥 = ﷮﷮ sec﷮2﷯ 2𝑥 – 3﷯−1﷯ ﷯𝑑𝑥 = ﷮﷮ sec﷮2﷯ 2𝑥 – 3﷯ ﷯𝑑𝑥− ﷮﷮ 1﷯.𝑑𝑥 = ﷮﷮ sec﷮2﷯ 2𝑥 – 3﷯ ﷯𝑑𝑥 − 𝑥+𝐶1 Solving 𝐈1 I1 = ﷮﷮ sec﷮2﷯ 2𝑥 – 3﷯ ﷯𝑑𝑥 Let 2𝑥 – 3=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 2−0 = 𝑑𝑡﷮𝑑𝑥﷯ 2= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥 = 𝑑𝑡﷮2﷯ Thus, our equation becomes ∴ ﷮﷮ sec﷮2﷯ 2𝑥 – 3﷯ ﷯𝑑𝑥 = ﷮﷮ sec﷮2 ﷯𝑡 ﷯. 𝑑𝑡﷮2﷯ = 1﷮2﷯ ﷮﷮ sec﷮2 ﷯𝑡 ﷯.𝑑𝑡 = 1﷮2﷯ tan﷮𝑡﷯+𝐶2 = 1﷮2﷯ tan﷮ 2𝑥−3﷯﷯+ 𝐶2 Now, I = ﷮﷮ sec﷮2﷯ 2𝑥 – 3﷯ ﷯𝑑𝑥−𝑥+𝐶1 = I1 − 𝑥+𝐶1 = 1﷮2﷯ tan﷮ 2𝑥−3﷯﷯+ 𝐶2 −𝑥+𝐶1 = 𝟏﷮𝟐﷯ 𝒕𝒂𝒏﷮ 𝟐𝒙−𝟑﷯﷯ −𝒙+𝑪