Ex 7.2, 21 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
Ex 7.2
Ex 7.2, 1
Ex 7.2, 2
Ex 7.2, 3 Important
Ex 7.2, 4
Ex 7.2, 5 Important
Ex 7.2, 6
Ex 7.2, 7 Important
Ex 7.2, 8
Ex 7.2, 9
Ex 7.2, 10 Important
Ex 7.2, 11 Important
Ex 7.2, 12
Ex 7.2, 13
Ex 7.2, 14 Important
Ex 7.2, 15
Ex 7.2, 16
Ex 7.2, 17
Ex 7.2, 18
Ex 7.2, 19 Important
Ex 7.2, 20 Important
Ex 7.2, 21 You are here
Ex 7.2, 22 Important
Ex 7.2, 23
Ex 7.2, 24
Ex 7.2, 25
Ex 7.2, 26 Important
Ex 7.2, 27
Ex 7.2, 28
Ex 7.2, 29 Important
Ex 7.2, 30
Ex 7.2, 31
Ex 7.2, 32 Important
Ex 7.2, 33 Important
Ex 7.2, 34 Important
Ex 7.2, 35
Ex 7.2, 36 Important
Ex 7.2, 37
Ex 7.2, 38 (MCQ) Important
Ex 7.2, 39 (MCQ) Important
Transcript
Ex 7.2, 21 tan2 (2𝑥 – 3) Let I = tan2 (2𝑥 – 3) . 𝑑𝑥 = sec2 2𝑥 – 3−1 𝑑𝑥 = sec2 2𝑥 – 3 𝑑𝑥− 1.𝑑𝑥 = sec2 2𝑥 – 3 𝑑𝑥 − 𝑥+𝐶1 Solving 𝐈1 I1 = sec2 2𝑥 – 3 𝑑𝑥 Let 2𝑥 – 3=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 2−0 = 𝑑𝑡𝑑𝑥 2= 𝑑𝑡𝑑𝑥 𝑑𝑥 = 𝑑𝑡2 Thus, our equation becomes ∴ sec2 2𝑥 – 3 𝑑𝑥 = sec2 𝑡 . 𝑑𝑡2 = 12 sec2 𝑡 .𝑑𝑡 = 12 tan𝑡+𝐶2 = 12 tan 2𝑥−3+ 𝐶2 Now, I = sec2 2𝑥 – 3 𝑑𝑥−𝑥+𝐶1 = I1 − 𝑥+𝐶1 = 12 tan 2𝑥−3+ 𝐶2 −𝑥+𝐶1 = 𝟏𝟐 𝒕𝒂𝒏 𝟐𝒙−𝟑 −𝒙+𝑪
