Example 15 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
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Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii) Important
Example 7 (i)
Example 7 (ii) Important
Example 7 (iii)
Example 8 (i)
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Example 9 (iii) Important
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Example 15 Important You are here
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Example 21 Important
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Example 24
Example 25 Important Deleted for CBSE Board 2023 Exams
Example 26 Deleted for CBSE Board 2023 Exams
Example 27 (i)
Example 27 (ii) Important
Example 27 (iii)
Example 27 (iv) Important
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Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 15 Find ∫1▒((3 sinϕ −2) cosϕ )/(5 − cos^2ϕ − 4 sinϕ ) 𝑑ϕ Let 𝑡=sinϕ Differentiating w.r.t. ϕ 𝑑𝑡/𝑑ϕ=cosϕ 𝑑𝑡/cosϕ =𝑑ϕ Now we can write ∫1▒((3 sinϕ −2) cosϕ )/(5 − cos^2ϕ − 4 sinϕ ) 𝑑ϕ =∫1▒((3 sinϕ − 2) cosϕ )/(5 − (1 − sin^2ϕ) − 4 sinϕ ) 𝑑ϕ =∫1▒((3𝑡 − 2) cosϕ )/(5 − 1 + 𝑡^2 − 4𝑡) 𝑑𝑡/cosϕ =∫1▒((3𝑡 − 2) 𝑑𝑡 )/(4 + 𝑡^2 − 4𝑡) =∫1▒((3𝑡 − 2) 𝑑𝑡 )/(𝑡^2 + 2^2 − 2.2 𝑡) =∫1▒((3𝑡 − 2) 𝑑𝑡 )/(𝑡 − 2)^2 We can write integrand ((3𝑡 − 2))/(𝑡 − 2)^2 =(𝐴 )/(𝑡 − 2) + (𝐵 )/(𝑡 − 2)^2 〖𝑠𝑖𝑛〗^2ϕ+〖𝑐𝑜𝑠〗^2ϕ=1 〖𝑐𝑜𝑠〗^2 ϕ=1−〖𝑠𝑖𝑛〗^2 ϕ (3𝑡 − 2)/(𝑡 − 2)^2 =(𝐴(𝑡 − 2) + 𝐵)/(𝑡 − 2)^2 Cancelling denominator (3𝑡−2)=𝐴(𝑡−2)+𝐵 Putting t = 2 (3(2) − 2) = A (2 − 2) + B 6 − 2 = A × 0 + B 4 = B B = 4 Putting t = 0 (3(0) − 2) = A (0 − 2) + B −2 = −2A + B Putting B = 4 −2 = −2A + 4 −6 = −2A A = 3 Hence, we can write it as ((3𝑡 − 2))/(𝑡 − 2)^2 =(𝐴 )/(𝑡 − 2) + (𝐵 )/(𝑡 − 2)^2 ((3𝑡 − 2))/(𝑡 − 2)^2 = 3/(𝑡 − 2) + 4/(𝑡 − 2)^2 Now, our equation becomes ∫1▒(3𝑡 − 2)/(𝑡 − 2)^2 𝑑𝑡=∫1▒3/(𝑡 − 2) 𝑑𝑡+∫1▒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|sinϕ−2|−4/(sinϕ − 2) +𝐶 =3 𝑙𝑜𝑔〖(2−𝑠𝑖𝑛𝜙)〗−4/(𝑠𝑖𝑛𝜙 − 2) +𝐶 =3 𝑙𝑜𝑔〖(2−𝑠𝑖𝑛𝜙)〗−4/(−(2 − 𝑠𝑖𝑛𝜙)) +𝐶 =𝟑 𝒍𝒐𝒈〖(𝟐−𝒔𝒊𝒏𝝓)〗+𝟒/(𝟐 − 𝒔𝒊𝒏𝝓 ) +𝑪 Since 𝑠𝑖𝑛ϕ∈[−1 , 1] 𝑠𝑖𝑛ϕ<2 2−𝑠𝑖𝑛ϕ always positive |𝑠𝑖𝑛ϕ−2|=2−𝑠𝑖𝑛ϕ
