Ex 7.1, 5 Find anti derivative of sin⁡2𝑥 – 4𝑒3𝑥 Subtracting (1) & (2) sin⁡2𝑥−〖4𝑒〗^3𝑥=(−1)/2 (cos⁡2𝑥 )^′− 4/3 (𝑒^3𝑥 )^′ We know that (cos⁡2𝑥 )^′=−2 sin⁡2𝑥 (−1)/2 (cos⁡2𝑥 )^′=sin⁡2𝑥 𝐬𝐢𝐧 𝟐𝒙=(−𝟏)/𝟐 (𝒄𝒐𝒔⁡𝟐𝒙 )^′ We know that (𝑒^3𝑥 )^′=𝑒^3𝑥 . 3 ⇒ 〖1(𝑒^3𝑥 )〗^′/3=𝑒^3𝑥 ⇒𝑒^3𝑥=〖1(𝑒^3𝑥 )〗^′/3 ⇒〖𝟒𝒆〗^𝟑𝒙=𝟒/𝟑 (𝒆^𝟑𝒙 )^′ …(2) =(−1/2 cos⁡2𝑥−4/3 𝑒^3𝑥 )^′ ∴ Anti derivative is (−𝟏)/𝟐 𝒄𝒐𝒔⁡𝟐𝒙− 𝟒/𝟑 𝒆^𝟑𝒙

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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