Ex 7.2, 32  - Integrate 1 / 1 + cotx - Chapter 7 - Ex 7.2

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.2, 32 1﷮1+ cot﷮𝑥﷯﷯ Step 1: Simplify the given function ﷮﷮ 1﷮1 + cot﷮𝑥﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 1﷮1 + cos﷮𝑥﷯﷮ sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 1﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷮ sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ 𝑑𝑥 Multiplying & dividing by 2 = ﷮﷮ 2 sin﷮𝑥﷯﷮2 sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 Adding & subtracting 𝑐𝑜𝑠⁡𝑥 in numerator = ﷮﷮ sin﷮𝑥﷯ + sin﷮𝑥﷯ + cos﷮𝑥﷯ − cos﷮𝑥﷯﷮2 sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯ + sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯+ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯ ﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ 1+ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯ ﷯﷯ 𝑑𝑥 ∴ We get ﷮﷮ 1﷮1 + cot﷮𝑥﷯﷯﷯ 𝑑𝑥= 1﷮2﷯ ﷮﷮ 1+ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯ ﷯﷯𝑑𝑥 = 1﷮2﷯ 𝑥+ ﷮﷮ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯ ﷯﷯𝑑𝑥﷯ + 𝐶1 Solving 𝐈1 I1 = ﷮﷮ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ 𝑑𝑥 Let sin﷮𝑥﷯ + cos﷮𝑥﷯=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 cos﷮𝑥﷯−sin﷮𝑥﷯= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯ 𝑑𝑥= 𝑑𝑡﷮− sin﷮𝑥﷯ − cos﷮𝑥﷯﷯﷯ Thus, our equation becomes I1 = ﷮﷮ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ 𝑑𝑥 = ﷮﷮ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮𝑡﷯﷯ . 𝑑𝑡﷮− sin﷮𝑥﷯ − cos﷮𝑥﷯﷯﷯ = −1 ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯ = − log﷮ 𝑡﷯﷯+𝐶 = − log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+𝐶2 Putting the value of I1 in eq. (1) ∴ ﷮﷮ 1﷮1 + cot﷮𝑥﷯﷯ ﷯ = 1﷮2﷯ 𝑥+ ﷮﷮ sin﷮𝑥﷯ − cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯ ﷯﷯𝑑𝑥﷯ + 𝐶1 = 1﷮2﷯ 𝑥− log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+𝐶2 ﷯ +𝐶1 = 𝑥﷮2﷯− 1﷮2﷯ log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+𝐶1+ 𝐶2﷮2﷯ = 𝒙﷮𝟐﷯ − 𝟏﷮𝟐﷯ 𝒍𝒐𝒈﷮ 𝒔𝒊𝒏﷮𝒙﷯+ 𝒄𝒐𝒔﷮𝒙﷯﷯﷯+𝑪

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