Ex 7.6, 21 - Integrate e2x sin x - Chapter 7 NCERT - Integration by parts

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.6, 21 𝑒﷮2𝑥﷯ sin⁡𝑥 I = ﷮﷮ 𝑒﷮2𝑥﷯ sin⁡𝑥 .𝑑𝑥﷯ ∴ I = ﷮﷮ sin﷮𝑥﷯. 𝑒﷮2𝑥﷯ 𝑑𝑥﷯ I = sin﷮𝑥﷯ ﷮﷮ 𝑒﷮2𝑥﷯ 𝑑𝑥﷯﷯− ﷮﷮ 𝑑 sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ ﷮﷮ 𝑒﷮2𝑥﷯ 𝑑𝑥﷯﷯﷯𝑑𝑥 I = sin﷮𝑥﷯ . 𝑒﷮2𝑥﷯﷮2﷯ − ﷮﷮ cos﷮𝑥﷯ . 𝑒﷮2𝑥﷯﷮2﷯﷯𝑑𝑥 I = 1﷮2﷯ . 𝑒﷮2𝑥﷯ sin﷮𝑥﷯− 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ . 𝑒﷮2𝑥﷯﷯𝑑𝑥 Solving I2 = 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ . 𝑒﷮2𝑥﷯﷯𝑑𝑥 ∴ I1 = 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ . 𝑒﷮2𝑥﷯﷯𝑑𝑥 = 1﷮2﷯ cos﷮𝑥﷯ ﷮﷮ 𝑒﷮2𝑥﷯ 𝑑𝑥﷯− ﷮﷮ 𝑑 cos﷮𝑥﷯﷯﷮𝑑𝑥﷯ ﷮﷮ 𝑒﷮2𝑥﷯ 𝑑𝑥﷯﷯﷯𝑑𝑥﷯ = 1﷮2﷯ cos﷮𝑥﷯ . 𝑒﷮2𝑥﷯﷮2﷯ − ﷮﷮(− sin﷮𝑥﷯) . 𝑒﷮2𝑥﷯﷮2﷯﷯ 𝑑𝑥﷯ = 1﷮2﷯ 𝑒﷮2𝑥﷯ . cos﷮𝑥﷯ ﷮2﷯ + 1﷮2﷯ ﷮﷮ 𝑒﷮2𝑥﷯ sin﷮𝑥﷯﷯ 𝑑𝑥﷯ = 1﷮2﷯ 𝑒﷮2𝑥﷯ . cos﷮𝑥﷯ ﷮2﷯ + 1﷮2﷯ I﷯+𝐶1 Putting the value of I1 in (1) , we get I = ﷮﷮ 𝑒﷮2𝑥﷯ sin﷮𝑥﷯ 𝑑𝑥﷯ I = 𝑒﷮2𝑥﷯ sin﷮𝑥﷯﷮2﷯ − 1﷮2﷯ 𝑒﷮2𝑥﷯ . cos﷮𝑥﷯ ﷮2﷯ + I﷮2﷯﷯+𝐶1﷯ I = 𝑒﷮2𝑥﷯﷮2﷯ sin﷮𝑥﷯− 𝑒﷮2𝑥﷯ ﷮4﷯ cos﷮𝑥﷯− I﷮4﷯−𝐶1 𝐼+ I﷮4﷯ = 𝑒﷮2𝑥﷯ sin﷮𝑥﷯﷮2﷯ − 𝑒﷮2𝑥﷯ . cos﷮𝑥﷯ ﷮4﷯ 5𝐼﷮4﷯ = 𝑒﷮2𝑥﷯﷮4﷯ 2 sin﷮𝑥﷯− cos﷮𝑥﷯﷯− 𝐶1 𝐼 = 4﷮5﷯ . 𝑒﷮2𝑥﷯﷮4﷯ 2 sin﷮𝑥﷯− cos﷮𝑥﷯﷯− 4𝐶1﷮5﷯ 𝐼 = 𝒆﷮𝟐𝒙﷯﷮𝟓﷯ 𝟐 𝒔𝒊𝒏﷮𝒙﷯− 𝒄𝒐𝒔﷮𝒙﷯﷯+𝑪

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.