Ex 7.2, 21
Last updated at Dec. 8, 2016 by Teachoo
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
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