Ex 7.3, 1 - Chapter 7 Class 12 Integrals (Term 2)

Transcript

Ex 7.3, 1 Find the integral of sin2 (2𝑥 + 5) ∫1▒〖sin2 (2𝑥 + 5) 〗 𝑑𝑥 =∫1▒(1 − 〖𝑐𝑜𝑠 2〗⁡(2𝑥 + 5))/2 𝑑𝑥 =1/2 ∫1▒〖1−cos⁡(4𝑥+10) 〗 𝑑𝑥 =1/2 [∫1▒1 𝑑𝑥−∫1▒cos⁡(4𝑥+10) 𝑑𝑥] We know that cos 2𝜃=1−2 sin^2⁡𝜃 2 sin^2 𝜃=1−cos⁡2𝜃 sin^2 𝜃=1/2 [1−cos⁡2𝜃 ] Replace 𝜃 by (2𝑥+5) sin^2 (2𝑥+5)=(1 − cos⁡2(2𝑥 + 5))/2 (As ∫1▒cos⁡(𝑎𝑥+𝑏) 𝑑𝑥=sin⁡(𝑎𝑥 + 𝑏)/𝑎+𝐶) =1/2 [𝑥− sin⁡(4𝑥 + 10)/4 +𝐶] =𝒙/𝟐 − 𝟏/𝟖 𝒔𝒊𝒏⁡(𝟒𝒙+𝟏𝟎)+𝑪