Chapter 7 Class 12 Integrals
Concept wise

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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 26 Evaluate the definite integral ∫_0^(𝜋/2)▒〖(cos^2⁡𝑥 𝑑𝑥)/(cos^2⁡𝑥 + 4 sin^2⁡𝑥 ) 〗 Let I = ∫1_0^(𝜋/2)▒(〖𝑐𝑜𝑠〗^2 𝑥)/(〖𝑐𝑜𝑠〗^2 𝑥 + 4〖𝑠𝑖𝑛〗^2 𝑥) 𝑑𝑥 = ∫1_0^(𝜋/2)▒(〖𝑐𝑜𝑠〗^2 𝑥)/(〖𝑐𝑜𝑠〗^2 𝑥 + 4(〖1 − 𝑐𝑜𝑠〗^2 𝑥) 𝑑𝑥) = ∫1_0^(𝜋/2)▒(〖𝑐𝑜𝑠〗^2 𝑥)/(4 − 3 〖𝑐𝑜𝑠〗^2 𝑥) 𝑑𝑥 = (−1)/3 ∫1_0^(𝜋/4)▒〖 (−3 〖𝑐𝑜𝑠〗^2 𝑥 )/(4 − 3 〖𝑐𝑜𝑠〗^2 𝑥) 〗 𝑑𝑥 = (−1)/3 ∫1_0^(𝜋/2)▒(〖− 3 𝑐𝑜𝑠〗^2 𝑥 + 4 − 4)/(4 − 3 〖𝑐𝑜𝑠〗^2 𝑥) 𝑑𝑥 = (−1)/3 ∫1_0^(𝜋/2)▒(〖4 − 3 𝑐𝑜𝑠〗^2 𝑥 − 4)/(4 − 3 〖𝑐𝑜𝑠〗^2 𝑥) 𝑑𝑥 = (−1)/3 ∫1_0^(𝜋/2)▒〖1−4/(4 − 3 〖𝑐𝑜𝑠〗^2 𝑥)〗 𝑑𝑥 Dividing numerator and denominator by 〖𝑐𝑜𝑠〗^2 𝑥 = (−1)/3 (𝜋/2)+4/3 ∫1_0^(𝜋/2)▒(𝑑𝑥/(〖𝑐𝑜𝑠〗^2 𝑥))/((4 − 3 〖𝑐𝑜𝑠〗^2 𝑥 )/(〖𝑐𝑜𝑠〗^2 𝑥)) 𝑑𝑥 = (−1)/3 (𝜋/2)+4/3 ∫1_0^(𝜋/2)▒(〖𝑠𝑒𝑐〗^2 𝑥)/(4 〖𝑠𝑒𝑐〗^2 𝑥 − 3) 𝑑𝑥 = (−𝜋)/6+4/3 ∫1_0^(𝜋/2)▒(〖𝑠𝑒𝑐〗^2 𝑥)/(4 (1 + 〖𝑡𝑎𝑛〗^2 𝑥) − 3) 𝑑𝑥 = (−𝜋)/6+4/3 ∫1_0^(𝜋/2)▒(〖𝑠𝑒𝑐〗^2 𝑥)/(4 + 4 〖𝑡𝑎𝑛〗^2 𝑥 − 3) 𝑑𝑥 = (−𝜋)/6+4/3 ∫1_0^(𝜋/2)▒(〖𝑠𝑒𝑐〗^2 𝑥)/( 4 〖𝑡𝑎𝑛〗^2 𝑥 + 1) 𝑑𝑥 Let tan x = t Differentiating w.r.t x 〖𝑠𝑒𝑐〗^2 x dx = dt Thus, When x = 0, t = 0, & when x = 𝜋/2, 𝑡= ∞ Substituting values and limit I = (−𝜋)/6+4/3 ∫1_0^(𝜋/2)▒(〖𝑠𝑒𝑐〗^2 𝑥)/( 4 〖𝑡𝑎𝑛〗^2 𝑥 + 1) 𝑑𝑥 ∴ I =(−𝜋)/6+4/3 ∫1_0^∞▒𝑑𝑡/(〖4𝑡〗^2+1) = (−𝜋)/6+4/3 ∙1/4 ∫1_0^∞▒𝑑𝑡/(〖 𝑡〗^2+1/4) =(−𝜋)/6+4/3∙1/4 × 1/((1/2) ) [〖𝑡𝑎𝑛〗^(−1) 𝑡/(1/2)]_0^∞ = (−𝜋)/6+2/3∙ [〖𝑡𝑎𝑛〗^(−1) 2𝑡]_0^∞ =(−𝜋)/6+2/3∙〖[𝑡𝑎𝑛〗^(−1) ∞−〖𝑡𝑎𝑛〗^(−1) 0] = −𝜋/6+2/3∙[𝜋/2−0] =−𝜋/6+𝜋/3 =𝝅/𝟔

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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.