Ex 7.9, 22 - Direct Integrate dx / 4 + 9x2 from 0 to 2/3 - Definate Integration - By Formulae

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Ex7.9, 22 Choose the correct answer ∫_0^(2/3)▒𝑑π‘₯/(4 +9π‘₯^2 ) equals (A) πœ‹/6 (B) πœ‹/12 (C) πœ‹/24 (D) πœ‹/4 Step 1 :- Let F(π‘₯)=∫1▒𝑑π‘₯/(4 + 9π‘₯^2 ) Divide and multiply the integrate by 4 =∫1β–’β–ˆ(𝑑π‘₯@4)/((4 + 9π‘₯^2)/4) =1/4 ∫1▒𝑑π‘₯/(1 + 9/4 π‘₯^2 ) =1/4 ∫1▒𝑑π‘₯/(1 + (3/2 π‘₯)^2 ) Put 3/2 π‘₯=𝑑 Differentiating w.r.t.π‘₯ 𝑑(3/2 π‘₯)/𝑑π‘₯=𝑑𝑑/𝑑π‘₯ 3/2=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(3/2) 𝑑π‘₯=2/3 𝑑𝑑 Hence 1/4 ∫1▒〖𝑑π‘₯/(1+(3/2 π‘₯)^2 )=1/4 ∫1β–’γ€–1/(1+𝑑^2 ) 2/3γ€— 𝑑𝑑〗 =1/4 Γ— 2/3 ∫1▒𝑑𝑑/(1+𝑑^2 ) =1/6 tan^(βˆ’1)⁑𝑑 Putting 𝑑=3/2 π‘₯ =1/6 tan^(βˆ’1)⁑〖3/2 π‘₯γ€— Hence F(π‘₯)=1/6 tan^(βˆ’1)⁑〖3/2 π‘₯γ€— Step 2 :- ∫_0^(2/3)▒〖𝑑π‘₯/(4+9π‘₯^2 )=𝐹(2/3)βˆ’πΉ(0) γ€— =1/6 tan^(βˆ’1)⁑〖(3/2 Γ— 2/3)βˆ’1/6 tan^(βˆ’1)⁑(0) γ€— =1/6 tan^(βˆ’1)⁑〖(1)βˆ’1/6 tan^(βˆ’1)⁑(0) γ€— =1/6 Γ— πœ‹/4βˆ’0 =πœ‹/24 ∡ Option (C) is correct.

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