Ex 7.2, 21 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
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
Integration by substitution - Trignometric - Normal
Example 5 (i)
Ex 7.2, 22 Important
Misc 15
Example 37
Ex 7.2, 27
Ex 7.2, 31
Ex 7.2, 30
Ex 7.2, 26 Important
Ex 7.2, 24
Example 6 (i)
Ex 7.2, 29 Important
Ex 7.2, 21 You are here
Ex 7.2, 34 Important
Ex 7.2, 39 (MCQ) Important
Ex 7.2, 25
Ex 7.2, 32 Important
Ex 7.2, 33 Important
Misc 7 Important
Integration by substitution - Trignometric - Normal
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