# Ex 7.6, 21 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.6, 21 𝑒2𝑥 sin𝑥 I = 𝑒2𝑥 sin𝑥 .𝑑𝑥 ∴ I = sin𝑥. 𝑒2𝑥 𝑑𝑥 I = sin𝑥 𝑒2𝑥 𝑑𝑥− 𝑑 sin𝑥𝑑𝑥 𝑒2𝑥 𝑑𝑥𝑑𝑥 I = sin𝑥 . 𝑒2𝑥2 − cos𝑥 . 𝑒2𝑥2𝑑𝑥 I = 12 . 𝑒2𝑥 sin𝑥− 12 cos𝑥 . 𝑒2𝑥𝑑𝑥 Solving I2 = 12 cos𝑥 . 𝑒2𝑥𝑑𝑥 ∴ I1 = 12 cos𝑥 . 𝑒2𝑥𝑑𝑥 = 12 cos𝑥 𝑒2𝑥 𝑑𝑥− 𝑑 cos𝑥𝑑𝑥 𝑒2𝑥 𝑑𝑥𝑑𝑥 = 12 cos𝑥 . 𝑒2𝑥2 − (− sin𝑥) . 𝑒2𝑥2 𝑑𝑥 = 12 𝑒2𝑥 . cos𝑥 2 + 12 𝑒2𝑥 sin𝑥 𝑑𝑥 = 12 𝑒2𝑥 . cos𝑥 2 + 12 I+𝐶1 Putting the value of I1 in (1) , we get I = 𝑒2𝑥 sin𝑥 𝑑𝑥 I = 𝑒2𝑥 sin𝑥2 − 12 𝑒2𝑥 . cos𝑥 2 + I2+𝐶1 I = 𝑒2𝑥2 sin𝑥− 𝑒2𝑥 4 cos𝑥− I4−𝐶1 𝐼+ I4 = 𝑒2𝑥 sin𝑥2 − 𝑒2𝑥 . cos𝑥 4 5𝐼4 = 𝑒2𝑥4 2 sin𝑥− cos𝑥− 𝐶1 𝐼 = 45 . 𝑒2𝑥4 2 sin𝑥− cos𝑥− 4𝐶15 𝐼 = 𝒆𝟐𝒙𝟓 𝟐 𝒔𝒊𝒏𝒙− 𝒄𝒐𝒔𝒙+𝑪

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Ex 7.6, 21 You are here

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About the Author

Davneet Singh

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.