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Example 15 - Find (3 sin - 2) cos / 5 - cos2 - 4 sin - Integration by partial fraction - Type 2

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Example 15 Find ﷮﷮ 3 sin﷮ϕ﷯ −2﷯ cos﷮ϕ﷯ ﷮5 − cos﷮2﷯﷮ϕ﷯ − 4 sin﷮ϕ﷯﷯﷯ 𝑑ϕ Let 𝑡= sin﷮ϕ﷯ Differentiating w.r.t. ϕ 𝑑𝑡﷮𝑑ϕ﷯= cos﷮ϕ﷯ 𝑑𝑡﷮ cos﷮ϕ﷯﷯=𝑑ϕ Now we can write ﷮﷮ 3 sin﷮ϕ﷯ −2﷯ cos﷮ϕ﷯ ﷮5 − cos﷮2﷯﷮ϕ﷯ − 4 sin﷮ϕ﷯﷯﷯ 𝑑ϕ= ﷮﷮ 3 sin﷮ϕ﷯ −2﷯ cos﷮ϕ﷯ ﷮5 − sin﷮2﷯﷮ϕ﷯ − 1 − 4 sin﷮ϕ﷯﷯﷯ 𝑑ϕ = ﷮﷮ 3𝑡 −2﷯ cos﷮ϕ﷯ ﷮5 − 1 + 𝑡﷮2﷯ − 4𝑡﷯﷯ 𝑑𝑡﷮ cos﷮ϕ﷯﷯ = ﷮﷮ 3𝑡 −2﷯ 𝑑𝑡 ﷮4 + 𝑡﷮2﷯ − 4𝑡﷯﷯ = ﷮﷮ 3𝑡 − 2﷯ 𝑑𝑡 ﷮ 𝑡﷮2﷯ + 2﷮2﷯ − 2.2 𝑡﷯﷯ = ﷮﷮ 3𝑡 − 2﷯ 𝑑𝑡 ﷮ 𝑡 − 2﷯﷮2﷯﷯﷯ We can write integrand 3𝑡 − 2﷯﷮ 𝑡 − 2﷯﷮2﷯﷯= 𝐴 ﷮𝑡 − 2﷯ + 𝐵 ﷮ 𝑡 − 2﷯﷮2﷯﷯ 3𝑡 − 2﷮ 𝑡 − 2﷯﷮2﷯﷯= 𝐴 𝑡 − 2﷯ + 𝐵﷮ 𝑡 − 2﷯﷮2﷯﷯ By Cancelling denominator 3𝑡−2﷯=𝐴 𝑡−2﷯+𝐵 3𝑡−2﷯=𝐴 𝑡−2﷯+𝐵 Hence we can write it as 3𝑡 − 2﷯﷮ 𝑡 − 2﷯﷮2﷯﷯ = 3﷮𝑡 − 2﷯ + 4﷮ 𝑡 − 2﷯﷮2﷯﷯ Now, our equation becomes ﷮﷮ 3𝑡 − 2﷮ 𝑡 − 2﷯﷮2﷯﷯﷯ 𝑑𝑡= ﷮﷮ 3﷮𝑡 − 2﷯﷯𝑑𝑡+ ﷮﷮ 4﷮ 𝑡 − 2﷯﷮2﷯﷯﷯𝑑𝑡 =3 log﷮ 𝑡−2﷯﷯+4× 𝑡 − 2﷯﷮−1﷯﷮−1﷯ +𝐶 =3 log﷮ 𝑡−2﷯﷯−4× 1﷮ 𝑡 − 2﷯﷯ +𝐶 Substituting back the value of t =3 log﷮ sin﷮ϕ﷯−2﷯﷯− 4﷮ sin﷮ϕ﷯ − 2﷯ +𝐶 =𝟑 𝒍𝒐𝒈﷮𝟐− 𝒔𝒊𝒏﷮𝝓﷯﷯+ 𝟒﷮𝟐 − 𝒔𝒊𝒏﷮𝝓﷯﷯ +𝑪

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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