Ex 7.6, 10 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.6, 10 (sin^(−1)⁡𝑥 )^2 ∫1▒(sin^(−1)⁡𝑥 )^2 𝑑𝑥 Let sin^(−1)⁡𝑥=𝜃 ∴ 𝑥=sin⁡𝜃 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝜃=cos⁡𝜃 𝑑𝑥=cos⁡𝜃.𝑑𝜃 Thus, our equation becomes ∫1▒(sin^(−1)⁡𝑥 )^2 𝑑𝑥 = ∫1▒𝜃^2 . cos⁡𝜃.𝑑𝜃 =𝜃^2 ∫1▒〖cos⁡𝜃.𝑑𝜃〗−∫1▒(𝑑(𝜃^2 )/𝑑𝜃 ∫1▒〖cos⁡𝜃.𝑑𝜃〗)𝑑𝜃 =𝜃^2 sin⁡𝜃−∫1▒〖2𝜃.sin⁡𝜃. 𝑑𝜃〗 =𝜃^2 sin⁡𝜃−2∫1▒〖𝜽 𝒔𝒊𝒏⁡𝜽. 𝒅𝜽〗 We know that ∫1▒〖𝑓(𝑥) 𝑔⁡(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓^′ (𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = θ2 and g(x) = cos θ Solving I1 ∫1▒〖𝜽 𝒔𝒊𝒏⁡𝜽. 𝒅𝜽〗 = 𝜃∫1▒〖sin⁡𝜃 𝑑𝜃〗−∫1▒(𝑑𝜃/𝑑𝜃 ∫1▒〖sin⁡𝜃 𝑑𝜃〗) 𝑑𝜃 = 𝜃(−cos⁡𝜃 )−∫1▒〖1.(−cos⁡𝜃 ) 〗 𝑑𝜃 = −𝜃 cos⁡𝜃+∫1▒cos⁡𝜃 𝑑𝜃 = (−𝜃 cos⁡𝜃+sin⁡𝜃 )+𝐶1 Now we know that ∫1▒〖𝑓(𝑥) 𝑔⁡(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓′(𝑥)∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = θ and g(x) = sin θ Putting value of I1 in our equation ∫1▒(sin^(−1)⁡𝑥 )^2 𝑑𝑥 = 𝜃^2 sin⁡𝜃−2∫1▒〖𝜽 𝒔𝒊𝒏⁡𝜽. 𝒅𝜽〗 =𝜃^2 sin⁡𝜃−2(−𝜃 cos⁡𝜃+sin⁡𝜃+𝐶1) =𝜃^2 sin⁡𝜃+2𝜃 cos⁡𝜃−2 sin⁡𝜃−2𝐶1 =𝜃^2 sin⁡𝜃+2𝜃√(1−sin^2⁡〖𝜃 〗 )−2 sin⁡𝜃−𝑪𝟏 =𝜃^2 sin⁡𝜃+2𝜃√(1−sin^2⁡〖𝜃 〗 )−2 sin⁡𝜃+𝑪 =(〖𝐬𝐢𝐧〗^(−𝟏)⁡𝒙 )^𝟐 𝒙+𝟐(〖𝐬𝐢𝐧〗^(−𝟏)⁡𝒙 ) √(〖𝟏−〗⁡〖𝒙^𝟐 〗 )−𝟐𝒙+𝑪 𝑈𝑠𝑖𝑛𝑔 𝜃=sin^(−1)⁡𝑥 & sin⁡𝜃=𝑥