Evaluate the following  Definite Integral ∫ (-π/4) to (π/4)  (x + π/4)/(2 - cos ⁑2x )  dx

OR

Evaluate Definite Integral ∫ -2 to 2 (3x - 2x + 4) as the limit of a sum.

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Question 27 Evaluate the following ∫1_((βˆ’πœ‹)/4)^(πœ‹/4)β–’(π‘₯+πœ‹/4)/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ ∫1_((βˆ’πœ‹)/4)^(πœ‹/4)β–’(π‘₯+πœ‹/4)/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ ∫1_((βˆ’πœ‹)/4)^(πœ‹/4)β–’π‘₯/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ ∫1_((βˆ’πœ‹)/4)^(πœ‹/4)β–’(πœ‹/4)/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ Let f(x) = π‘₯/(2 βˆ’ cos⁑2π‘₯ ) f(–x) = (βˆ’π‘₯)/(2 βˆ’ cos⁑〖(βˆ’2π‘₯)γ€— ) As cos (–x) = cos x f(–x) = (βˆ’π‘₯)/(2 βˆ’ cos⁑π‘₯ ) ∴ f(–x) = – f(x) Let f(x) = (πœ‹/4)/(2 βˆ’ cos⁑2π‘₯ ) f(–x) = (πœ‹/4)/(2 βˆ’ cos⁑〖(βˆ’2π‘₯)γ€— ) As cos (–x) = cos x f(–x) = (πœ‹/4)/(2 βˆ’ cos⁑π‘₯ ) ∴ f(–x) = f(x) Using property If f(-x) = – f(x) ∫_(βˆ’π‘Ž)^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯γ€—=0 Using property If f(-x) = f(x) ∫_(βˆ’π‘Ž)^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯γ€—=2∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯ ∫1_((βˆ’πœ‹)/4)^(πœ‹/4)β–’π‘₯/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ ∫1_((βˆ’πœ‹)/4)^(πœ‹/4)β–’(πœ‹/4)/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ Thus, I = I1 + I2 I ∫1_0^(πœ‹/4)β–’(πœ‹/4)/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ ∫1_0^(πœ‹/4)β–’1/(2 βˆ’cos⁑2π‘₯ ) 𝑑π‘₯ ∫1_0^(πœ‹/4)β–’1/(2 βˆ’(1βˆ’2 sin^2⁑π‘₯) ) 𝑑π‘₯ ∫1_0^(πœ‹/4)β–’1/(1+2 sin^2⁑π‘₯ ) 𝑑π‘₯ ∫1_0^(πœ‹/4)β–’(1/cos^2⁑π‘₯ )/((1+2 sin^2⁑π‘₯)/cos^2⁑π‘₯ ) 𝑑π‘₯ ∫1_0^(πœ‹/4)β–’sec^2⁑π‘₯/(1+tan^2⁑π‘₯+2 tan^2⁑π‘₯ ) 𝑑π‘₯ Let tan x = t sec2 x dx = dt As x = 0 t = tan 0 = 0 As x = πœ‹/4, t = tan πœ‹/4 = 1 ∫1_0^1▒𝑑𝑑/3(1/3+t^2 ) ∫1_0^1▒𝑑𝑑/((1/√3)^2+t^2 ) Using ∫1▒〖𝑑π‘₯/(π‘Ž^2 + π‘₯^2 )=γ€–1/π‘Ž tan^(βˆ’1)〗⁑〖π‘₯/π‘Žγ€— γ€— = πœ‹/6 [γ€–1/((1/√3) ) tan^(βˆ’1)〗⁑〖𝑑/((1/√3) )γ€— ]_0^1 = (√3 πœ‹)/6 [tan^(βˆ’1)β‘γ€–βˆš3(1)γ€—βˆ’tan^(βˆ’1)β‘γ€–βˆš3(0)γ€— ] = (√3 πœ‹)/6 [tan^(βˆ’1)⁑√3βˆ’tan^(βˆ’1)⁑0 ] = (√3 πœ‹)/6 [πœ‹/3βˆ’0] = (√3 πœ‹)/6 [πœ‹/3] = (βˆšπŸ‘ 𝝅^𝟐)/πŸπŸ– Question 27 Evaluate as the limit of a sum. ∫1_(βˆ’2)^2β–’γ€–(3π‘₯^2βˆ’2π‘₯+4)γ€— We know that ∫1▒𝑓(π‘₯) 𝑑π‘₯ =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Putting π‘Ž =βˆ’2 𝑏 =2 β„Ž=(2 βˆ’ (βˆ’2))/𝑛 =(2 + 2)/𝑛 =4/𝑛 Now, 𝑓(π‘₯)=3π‘₯^2βˆ’2π‘₯+4 𝑓(βˆ’2)=3(βˆ’2)^2βˆ’2(βˆ’2)+4=3(4)+4+4=20 𝑓(βˆ’2+β„Ž)=3γ€–(βˆ’2+β„Ž)γ€—^2βˆ’2(βˆ’2+β„Ž)+4 =3(4+β„Ž^2βˆ’4β„Ž)+4βˆ’2β„Ž+4 =3β„Ž^2βˆ’14β„Ž+20 𝑓 (βˆ’2+2β„Ž)=3γ€–(βˆ’2+2β„Ž)γ€—^2βˆ’2(βˆ’2+2β„Ž)+4 =3(4+γ€–4β„Žγ€—^2βˆ’8β„Ž)+4βˆ’4β„Ž+4 " "=12β„Ž^2βˆ’28β„Ž+20 𝑓 (βˆ’2+3β„Ž)=3γ€–(βˆ’2+3β„Ž)γ€—^2βˆ’2(βˆ’2+3β„Ž)+4 =3(4+γ€–9β„Žγ€—^2βˆ’12β„Ž)+4βˆ’6β„Ž+4 " "=27β„Ž^2βˆ’42β„Ž+20 …. …… 𝑓(βˆ’2+(π‘›βˆ’1)β„Ž)=3γ€–(βˆ’2+(π‘›βˆ’1)β„Ž)γ€—^2βˆ’2(βˆ’2+(π‘›βˆ’1)β„Ž)+4 =3(4+γ€–(π‘›βˆ’1)^2 β„Žγ€—^2βˆ’4(π‘›βˆ’1)β„Ž)+4βˆ’2(π‘›βˆ’1)β„Ž+4 " "=γ€–3(π‘›βˆ’1)γ€—^2 β„Ž^2βˆ’14(π‘›βˆ’1)β„Ž+20 Hence we can write it as =(2βˆ’(βˆ’2)) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 20+(3β„Ž^2βˆ’14β„Ž+20)+(12β„Ž^2βˆ’28β„Ž+20)+ 20+(3β„Ž^2βˆ’14β„Ž+20)+(12β„Ž^2βˆ’28β„Ž+20)+ (27β„Ž^2βˆ’42β„Ž+20) + … …. +(γ€–3(π‘›βˆ’1)γ€—^2 β„Ž^2βˆ’14(π‘›βˆ’1)β„Ž+20) + (3β„Ž^2+12β„Ž^2+27β„Ž^2+ ………(3(π‘›βˆ’1)^2 β„Ž^2 ) – (14β„Ž+28β„Ž+42β„Ž+ …..(π‘›βˆ’1)β„Ž ) + 3h^2 (1+4+9+ ……+(π‘›βˆ’1)^2 ) – 14h(1+2+3+ ……..+(π‘›βˆ’1)) + 3h^2 (1^2+2^2+3^3+ ……+(π‘›βˆ’1)^2 ) – 14h(1+2+3+ ……..+(π‘›βˆ’1)) We know that 1^2+2^2+ …+𝑛^2= (𝑛 (𝑛 + 1)(2𝑛 + 1))/6 1^2+2^2+ ……+(π‘›βˆ’1)^2 = ((𝑛 βˆ’ 1) (𝑛 βˆ’1 + 1)(2(𝑛 βˆ’ 1) + 1))/6 = ((𝑛 βˆ’ 1) 𝑛 (2𝑛 βˆ’ 2 + 1) )/6 = (𝑛 (𝑛 βˆ’ 1) (2𝑛 βˆ’ 1) )/6 We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+(π‘›βˆ’1) = ((𝑛 βˆ’ 1) (𝑛 βˆ’1 + 1))/2 = (𝑛 (𝑛 βˆ’ 1) )/2 𝑛+3h^2 Γ—(𝑛(𝑛 βˆ’ 1)(2𝑛 βˆ’ 1))/6βˆ’14β„Ž Γ—(𝑛(𝑛 βˆ’ 1))/2 Putting h = 4/𝑛 =4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 =4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 =4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) =4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) =4 =4 =4 =4(20+16βˆ’28) =4 Γ—8 =πŸ‘πŸ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
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