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Example 15 - Find (3 sin - 2) cos / 5 - cos2 - 4 sin - Examples

Example 15 - Chapter 7 Class 12 Integrals - Part 2
Example 15 - Chapter 7 Class 12 Integrals - Part 3 Example 15 - Chapter 7 Class 12 Integrals - Part 4 Example 15 - Chapter 7 Class 12 Integrals - Part 5


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Example 15 Find ∫1▒((3 sin⁡ϕ −2) cos⁡ϕ )/(5 − cos^2⁡ϕ − 4 sin⁡ϕ ) 𝑑ϕ Let 𝑡=sin⁡ϕ Differentiating w.r.t. ϕ 𝑑𝑡/𝑑ϕ=cos⁡ϕ 𝑑𝑡/cos⁡ϕ =𝑑ϕ Now we can write ∫1▒((3 sin⁡ϕ −2) cos⁡ϕ )/(5 − cos^2⁡ϕ − 4 sin⁡ϕ ) 𝑑ϕ =∫1▒((3 sin⁡ϕ − 2) cos⁡ϕ )/(5 − (1 − sin^2⁡ϕ) − 4 sin⁡ϕ ) 𝑑ϕ =∫1▒((3𝑡 − 2) cos⁡ϕ )/(5 − 1 + 𝑡^2 − 4𝑡) 𝑑𝑡/cos⁡ϕ =∫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〗^2⁡ϕ+〖𝑐𝑜𝑠〗^2⁡ϕ=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⁡|sin⁡ϕ−2|−4/(sin⁡ϕ − 2) +𝐶 =3 𝑙𝑜𝑔⁡〖(2−𝑠𝑖𝑛⁡𝜙)〗−4/(𝑠𝑖𝑛⁡𝜙 − 2) +𝐶 =3 𝑙𝑜𝑔⁡〖(2−𝑠𝑖𝑛⁡𝜙)〗−4/(−(2 − 𝑠𝑖𝑛⁡𝜙)) +𝐶 =𝟑 𝒍𝒐𝒈⁡〖(𝟐−𝒔𝒊𝒏⁡𝝓)〗+𝟒/(𝟐 − 𝒔𝒊𝒏⁡𝝓 ) +𝑪 Since 𝑠𝑖𝑛⁡ϕ∈[−1 , 1] 𝑠𝑖𝑛⁡ϕ<2 2−𝑠𝑖𝑛⁡ϕ always positive |𝑠𝑖𝑛⁡ϕ−2|=2−𝑠𝑖𝑛⁡ϕ

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.