Ex 7.10, 2 - Evaluate using substitution: root sin cos 5 - Definate Integration - By Formulae

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex7.10, 2 Evaluate the integrals using substitution 0﷮ 𝜋﷮2﷯﷮ ﷮ sin﷮ ϕ ﷯﷯ cos﷮5﷯﷮ϕ﷯𝑑ϕ﷯ Let 𝐼= 0﷮ 𝜋﷮2﷯﷮ ﷮𝑠𝑖𝑛ϕ﷯ cos﷮5﷯﷮ϕ 𝑑ϕ﷯﷯ 𝐼= 0﷮ 𝜋﷮2﷯﷮ ﷮𝑠𝑖𝑛ϕ﷯ cos﷮4﷯﷮ϕ 𝑐𝑜𝑠ϕ 𝑑ϕ﷯﷯ 𝐼= ﷮﷮ ﷮ sin﷮ϕ﷯﷯ 1− sin﷮2﷯﷮ϕ﷯﷯﷮2﷯𝑐𝑜𝑠ϕ 𝑑ϕ﷯ Put 𝑡= sin﷮ϕ﷯ Differentiating w.r.t.ϕ 𝑑𝑡﷮𝑑ϕ﷯= 𝑑 sin﷮ϕ﷯﷯﷮𝑑ϕ﷯ 𝑑𝑡﷮𝑑ϕ﷯= cos﷮ϕ﷯ 𝑑𝑡﷮ cos﷮ϕ﷯﷯=𝑑ϕ Hence, when ϕ varies form 0 to 𝜋﷮2﷯ , then 𝑡 varies form 0 to 1 Hence we can write the integrate as 𝐼= 0﷮ 𝜋﷮2﷯﷮ ﷮𝑠𝑖𝑛ϕ﷯ 1− sin﷮2﷯﷮ϕ﷯﷯﷮2﷯𝑐𝑜𝑠ϕ 𝑑ϕ﷯ = 0﷮1﷮ ﷮𝑡﷯ 1− 𝑡﷮2﷯﷯﷮2﷯𝑐𝑜𝑠ϕ 𝑑𝑡﷮𝑐𝑜𝑠ϕ﷯﷯ = 0﷮1﷮ ﷮𝑡﷯ 1− 𝑡﷮2﷯﷯﷮2﷯𝑑𝑡﷯ = 0﷮1﷮ ﷮𝑡﷯ 1− 𝑡﷮2﷯﷯﷮2﷯𝑑𝑡﷯ = 0﷮1﷮ 𝑡﷮ 1﷮2﷯﷯ 1− 2𝑡﷮2﷯+ 𝑡﷮4﷯﷯𝑑𝑡﷯ = 0﷮1﷮ 𝑡﷮ 1﷮2﷯﷯− 2𝑡﷮2 + 1﷮2﷯﷯+ 𝑡﷮4 + 1﷮2﷯﷯﷯ 𝑑𝑡﷯ = 0﷮1﷮ 𝑡﷮ 1﷮2﷯﷯ 𝑑𝑡﷯−2 ﷮﷮ 𝑡﷮ 3﷮2﷯﷯﷯ 𝑑𝑡+ ﷮﷮ 𝑡﷮ 9﷮2﷯﷯﷯ 𝑑𝑡 = 𝑡﷮ 1﷮2﷯ +1﷯﷮ 1﷮2﷯ +1﷯﷯﷮0﷮1﷯−2 𝑡﷮ 3﷮2﷯ +1﷯﷮ 3﷮2﷯ +1﷯﷯﷮0﷮1﷯+ 𝑡﷮ 9﷮2﷯ +1﷯﷮ 9﷮2﷯ +1﷯﷯﷮0﷮1﷯ = 𝑡﷮ 3﷮2﷯ ﷯﷮ 3﷮2﷯﷯﷯﷮0﷮1﷯−2 𝑡﷮ 7﷮2﷯ ﷯﷮ 7﷮2﷯﷯﷯﷮0﷮1﷯+ 𝑡﷮ 11﷮2﷯ ﷯﷮ 11﷮2﷯﷯﷯﷮0﷮1﷯ = 2﷮3﷯ 1﷮ 3﷮2﷯﷯− 0﷮ 3﷮2﷯﷯﷯−2 × 2﷮7﷯ 1﷮ 7﷮2﷯﷯− 0﷮ 7﷮2﷯﷯﷯+ 2﷮11﷯ 1﷮ 11﷮2﷯﷯− 0﷮ 11﷮2﷯﷯﷯ = 2﷮3﷯− 4﷮7﷯+ 2﷮11﷯ = 154 − 132 + 42﷮231﷯ = 𝟔𝟒﷮𝟐𝟑𝟏﷯

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